100 Advanced SAT & Olympiad-Level Trigonometry Questions
Focusing on algebraic techniques and identities (no calculus).
Question 1
In a right triangle, angles \(A\) and \(B\) satisfy \(A + B = 90^\circ\).
Prove that \(\sin(A) = \cos(B)\) and \(\tan(A) = \cot(B)\).
Use complementary angles: \(A + B = 90^\circ\). Express one angle in terms of the other, and recall definitions of sine and cosine in a right triangle.
Since \(A + B = 90^\circ\), \(B = 90^\circ - A\). Then \(\sin(A) = \sin(A)\) but \(\cos(B) = \cos(90^\circ - A) = \sin(A)\), etc. This proves \(\tan(A) = \cot(B)\) as well.
Question 2
Convert \(\,210^\circ\) to radians and \(\,\frac{8\pi}{9}\) to degrees.
In a right triangle with acute angles of \(\,30^\circ\) and \(\,60^\circ\), show that the sides are in the ratio
1 : \(\sqrt{3}\) : 2.
Let the side opposite \(30^\circ\) be x, the side opposite \(60^\circ\) be y, and the hypotenuse be z. Use \(\sin(30^\circ) = \frac{1}{2}\), \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).
Prove the Pythagorean identity: \(\sin^2\theta + \cos^2\theta = 1\).
Think unit circle definitions of sine and cosine.
On the unit circle, \(\sin\theta = y\) and \(\cos\theta = x\) for a point \((x, y)\) on the circle \(x^2 + y^2 = 1\). Thus \(\sin^2\theta + \cos^2\theta = 1.\)
Question 7
Show that \(\csc(\theta) = \frac{1}{\sin(\theta)}\) and \(\sec(\theta) = \frac{1}{\cos(\theta)}\).
Use the definitions: csc is the reciprocal of sin, sec is the reciprocal of cos.
By definition, \(\csc(\theta) = \frac{1}{\sin(\theta)}\) and \(\sec(\theta) = \frac{1}{\cos(\theta)}\). These follow from reciprocal relationships of right-triangle sides or unit-circle definitions.
Question 8
A ladder 10 feet long leans against a wall, making an angle of \(\,60^\circ\) with the ground. Find the height it reaches on the wall.
Use \(\sin(60^\circ)\) or \(\cos(60^\circ)\), depending on which side you define as opposite or adjacent.
If \(y = \sin(x)\) has a maximum value of 1 at \(x = \frac{\pi}{2}\), find the coordinates of its peak in the \(xy\)-plane.
Evaluate \(\sin\bigl(\frac{\pi}{2}\bigr)\).
The peak is at \(\Bigl(\frac{\pi}{2}, 1\Bigr).\)
Question 10
A surveyor measures the angle of elevation to the top of a building as \(\,35^\circ\), standing 100 meters away from its base. Find the building's height (ignore eye-height).
Use \(\tan(35^\circ) = \frac{\text{opposite}}{\text{adjacent}}\).
Height \(= 100 \cdot \tan(35^\circ)\) meters (approximately \(70.0\) m if \(\tan(35^\circ)\approx 0.7002\)).
Question 11
Write the general form for the sine function: \(y = A \sin\bigl(B(x - C)\bigr) + D.\) Identify amplitude, period, phase shift, and vertical shift.
The function \(y = \cos(x)\) is shifted to the right by \(\frac{\pi}{3}\) and stretched vertically by 2. Write its new equation.
Horizontal shift inside the parentheses, amplitude change out front.
\(y = 2 \cos\bigl(x - \frac{\pi}{3}\bigr)\).
Question 13
Find the amplitude and period of \(y = -3 \sin(2x + \pi)\).
Amplitude \(= |A|\). Period \(= \frac{2\pi}{B}\).
Amplitude \(= 3;\) Period \(= \pi\) (since \(B = 2\)).
Question 14
Prove that if \(y = \sin^{-1}(x)\), then \(\sin(y) = x\). State the domain/range restrictions.
Inverse sine is the angle whose sine is \(x\).
By definition of inverse sine, \(y = \arcsin(x)\) means \(x = \sin(y)\). The domain of \(\arcsin(x)\) is \([-1,1]\), and the range is \(\bigl[-\frac{\pi}{2}, \frac{\pi}{2}\bigr]\).
Question 15
Use the Law of Sines to find angle \(A\) in a triangle with sides \(a=5\), \(b=7\), \(c=8\) and angle \(B\) opposite \(b\).
Law of Sines: \(\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\).
First find \(B\) using the Law of Cosines or Sines. For instance, by Law of Cosines:
\(\cos(B) = \frac{a^2 + c^2 - b^2}{2ac} = \frac{25 + 64 - 49}{2\cdot5\cdot8} = \frac{40}{80} = 0.5\) so \(B = 60^\circ\).
Then using \(\frac{a}{\sin(A)} = \frac{b}{\sin(B)}\) gives:
\(\frac{5}{\sin(A)} = \frac{7}{\sin(60^\circ)} = \frac{7}{\sqrt{3}/2} = \frac{14}{\sqrt{3}}.\)
Hence \(\sin(A) = 5 \cdot \frac{\sqrt{3}}{14} = \frac{5\sqrt{3}}{14}\) which is approximately \(0.6160\), so \(A \approx 19.47^\circ.\)
Question 16
A triangle has sides \(13, 14, 15\). Use the Law of Cosines to find its largest angle.
The largest angle is opposite the longest side, which is \(15\).
Let the largest angle be \(C\) opposite side \(15\). Then
\(\cos(C)= \frac{13^2 + 14^2 - 15^2}{2\cdot13\cdot14}
= \frac{169 + 196 - 225}{364}
= \frac{140}{364}
= \frac{35}{91}\approx 0.3846\)
so \(C\approx 67.5^\circ.\)
Question 17
Find the area of a triangle with sides \(7, 9\), and included angle \(\,60^\circ\) between them.
Solve the trigonometric equation: \(\sin(x) = \frac{\sqrt{3}}{2}\) for \(x\) in \([0, 2\pi]\).
\(\sin(x) = \frac{\sqrt{3}}{2}\) occurs at specific standard angles in the unit circle.
\(x = \frac{\pi}{3}\) or \(x = \frac{2\pi}{3}\) in \([0, 2\pi].\)
Question 19
Prove the identity: \(\cos(\alpha - \beta) = \cos\alpha\,\cos\beta + \sin\alpha\,\sin\beta.\)
Use sum formulas or unit circle geometry with coordinates for angles \(\alpha\) and \(\beta\).
By the standard angle addition formula, \(\cos(\alpha - \beta)= \cos\alpha\,\cos\beta + \sin\alpha\,\sin\beta\). It's derived from rotating vectors or expansions of \(e^{i\theta}\).
Question 20
Express \(\sin(75^\circ)\) using the sum of angles identity \(\sin\bigl(45^\circ + 30^\circ\bigr)\).
Solve the real-world problem: A pendulum swings such that its horizontal displacement is modeled by \(x(t)= 3\,\sin(2\pi\,t).\) Find how many oscillations occur in 5 seconds.
The period for \(\sin(k\,t)\) is \(\frac{2\pi}{k}.\)
The period is \(1\) second (since \(\frac{2\pi}{2\pi}=1\)).
In 5 seconds, there are 5 complete oscillations.
Question 27
In simple harmonic motion modeled by \(y(t)= 4\cos(3t)\), find the amplitude and angular frequency.
In \(y(t)= A\cos(\omega t)\), amplitude \(= A\), angular frequency \(= \omega.\)
Amplitude \(= 4,\) angular frequency \(= 3\) (radians per second).
Question 28
Solve the equation: \(2\,\sin^2(x) - 1 = 0\) in the interval \([0, 2\pi]\).
Set \(\sin^2(x)= \frac12\) => \(\sin(x)= \pm \frac{\sqrt{2}}{2}\).
Solve for \(x\) if \(\cos(x)= -\frac{\sqrt{3}}{2}\) and \(x\) in \([0, 2\pi]\).
\(\cos(x)= \frac{\sqrt{3}}{2}\) at \(x= \pm \frac{\pi}{6}\) from the principal angle. Negative means quadrant II or III.
\(x= \frac{5\pi}{6}, \frac{7\pi}{6}.\)
Question 30
Show that for an angle \(\theta\) in standard position, \(\tan(\theta)= \frac{y}{x}\), provided \(x\neq 0\), on the unit circle (scaled for radius \(r\)).
From \(\sin\theta= \frac{y}{r}\) and \(\cos\theta= \frac{x}{r}\) =>
\(\tan\theta= \frac{\sin\theta}{\cos\theta}= \frac{(y/r)}{(x/r)}= \frac{y}{x}.\)
Question 31
Find all solutions to \(3\,\tan(x)= \sqrt{3}\) in the interval \([0, 2\pi]\).
\(\tan(x)= \frac{\sqrt{3}}{3}= \frac{1}{\sqrt{3}}.\) This occurs at \(x= \frac{\pi}{6} + k\pi.\)
\(\tan(x)= \frac{1}{\sqrt{3}}\implies x= \frac{\pi}{6}, \frac{7\pi}{6}\) in \([0, 2\pi].\)
Question 32
Prove the identity: \(\sin(x) + \sin(y)= 2\,\sin\Bigl(\frac{x+y}{2}\Bigr)\,\cos\Bigl(\frac{x-y}{2}\Bigr).\)
Use sum-to-product formulas.
By sum-to-product, \(\sin(x)+ \sin(y)
= 2 \sin\Bigl(\frac{x+y}{2}\Bigr)\cos\Bigl(\frac{x-y}{2}\Bigr).\)
Question 33
Convert the rectangular coordinates \((-2, 2)\) to polar form \((r, \theta)\).
\(r= \sqrt{x^2 + y^2}, \theta= \arctan\bigl(\frac{y}{x}\bigr)\) with quadrant consideration.
\(r= \sqrt{(-2)^2 + 2^2} = \sqrt{4 +4}= \sqrt{8}= 2\sqrt{2}.\)
\(\theta= \arctan\Bigl(\frac{2}{-2}\Bigr)= \arctan(-1)= -\frac{\pi}{4},\) but the point is in quadrant II, so \(\theta= \frac{3\pi}{4}.\)
Question 34
Solve for \(x\) in \([0, 2\pi)\) if \(\sin(2x)= 1.\)
\(2x= \frac{\pi}{2} + 2k\pi\) or \(2x= \frac{3\pi}{2} +2k\pi.\) But \(\sin(\theta)=1\) only at \(\theta= \frac{\pi}{2}+ 2k\pi.\)
Evaluate \(\bigl(\cos 36^\circ + i \sin 36^\circ\bigr)^5\) using De Moivre’s theorem and simplify the argument.
\((r \,\mathrm{cis}(\theta))^n= r^n\,\mathrm{cis}(n\theta).\) Here \(r=1,\; n=5,\; \theta=36^\circ \implies n\theta=180^\circ.\)
\(\bigl(\cos 36^\circ + i \sin 36^\circ\bigr)^5= \cos(180^\circ) + i \sin(180^\circ)= -1 + 0i= -1.\)
Question 48
In the complex plane, find all solutions to \(z^3= -8.\)
\(-8= 8 \,\mathrm{cis}(\pi).\) Then \(z= 2\,\mathrm{cis}\Bigl(\frac{\pi+2k\pi}{3}\Bigr).\)
Let \(-8= 8\bigl(\cos\pi + i \sin\pi\bigr).\) Then
\(z= 2 \cos\Bigl(\frac{\pi+ 2k\pi}{3}\Bigr) + i \,\sin\Bigl(\frac{\pi+ 2k\pi}{3}\Bigr),\; k=0,1,2.\)
Thus 3 distinct roots.
Question 49
Convert the polar equation \(r= 3\) to rectangular coordinates.
\(r^2= x^2 + y^2 \implies\) if \(r=3,\) then \(r^2=9.\)
\(x^2 + y^2= 9,\) a circle radius 3 centered at \((0,0).\)
Question 50
Prove that \(\cos\bigl(\frac{\pi}{2} -\theta\bigr)= \sin(\theta)\) using co-function identities.
Use complementary angles or the unit circle approach.
\(\cos\Bigl(\frac{\pi}{2}- \theta\Bigr)= \sin(\theta).\) On the unit circle, shifting an angle by \(\frac{\pi}{2}\) swaps sine and cosine.
Question 51
Prove the identity: \(\tan\bigl(\theta + \frac{\pi}{4}\bigr)= \frac{\tan\theta+1}{1 - \tan\theta}.\)
Use \(\tan(\alpha+ \beta)= \frac{\tan\alpha+ \tan\beta}{1- \tan\alpha\,\tan\beta}\). Here \(\tan(\frac{\pi}{4})=1.\)
Solve the advanced equation: \(2\,\sin(2x)= \sqrt{3} + \sin(2x)\), for \(x\in [0, 2\pi].\)
Isolate \(\sin(2x)\).
\(2\,\sin(2x)= \sqrt{3} + \sin(2x) \implies \sin(2x)= \sqrt{3}.\)
But \(\sin(2x)\le1,\ \sqrt{3}\approx 1.732>1 \implies\) no real solution in \([0, 2\pi].\)
Question 54
Show that if \(\sin(x)= \cos(x)\), then \(x= \frac{\pi}{4}+ n\pi.\)
Show that \(\sin(A)+ \sin(B)+ \sin(C)= 4\,\cos\Bigl(\frac{A-B}{2}\Bigr) \cos\Bigl(\frac{B-C}{2}\Bigr)\,\sin\Bigl(\frac{A+C}{2}\Bigr)\) if \(A+B+C= \pi\) (angles of a triangle).
Use sum-to-product expansions and the fact that \(A+B+C= \pi \implies C= \pi- A- B.\)
A known trig identity for angles of a triangle.
One typical derivation uses sum-to-product on pairs of sines plus substituting \(C= \pi- A- B.\)
Question 60
Solve the system in terms of \(r, \theta\): \(x= r \cos\theta,\; y= r \sin\theta\) for \(x=2,\; y=2\sqrt{3}.\)
Given that \(\sin(\alpha)= \frac{4}{5}\), \(\alpha\) in quadrant I, and \(\cos(\beta)= \frac{12}{13}\), \(\beta\) in quadrant IV, find \(\sin(\alpha+ \beta).\)
Use \(\sin(\alpha+ \beta)= \sin\alpha \cos\beta + \cos\alpha \sin\beta.\)
If \(\cos(A)= \frac{4}{5}\) with \(A\) in quadrant IV, find \(\sin\bigl(\frac{A}{2}\bigr).\)
Use half-angle identity: \(\sin\bigl(\frac{A}{2}\bigr)= \pm \sqrt{\frac{1- \cos(A)}{2}}\). Determine sign from quadrant of \(\frac{A}{2}\).
Since \(A\) is in IV, \(\frac{A}{2}\) is in II (i.e. angle between \(\frac{\pi}{2}\) and \(\pi\)), so \(\sin(\frac{A}{2})\ge0.\)
\[
\sin\Bigl(\frac{A}{2}\Bigr)= \sqrt{\frac{1- \cos(A)}{2}}
= \sqrt{\frac{1- \frac{4}{5}}{2}}
= \sqrt{\frac{\frac{1}{5}}{2}}
= \sqrt{\frac{1}{10}}
= \frac{1}{\sqrt{10}}.
\]
Question 68
Let \(\tan(x)= 2.\) Find \(\sin(2x)\) in exact form, assuming \(x\) is in the first quadrant.
If \(\tan(x)= 2= \frac{2}{1},\) then opposite \(=2\), adjacent \(=1\) => hypotenuse \(= \sqrt{5}\) => \(\sin(x)= \frac{2}{\sqrt{5}}, \cos(x)= \frac{1}{\sqrt{5}}.\)
Show that the polar equation \(r= \theta\) is not a function in Cartesian form for all real \(\theta\) (explain why).
\(r= \theta\) in polar form spirals outward. Converting to \(x,y\) is complicated and not single-valued in \(x\) or \(y\).
The curve is a spiral passing multiple times over the same region, giving multiple \(y\)-values for a single \(x\). Hence it does not represent a single-valued function \(y(x)\).
Question 74
Suppose parametric equations \(x= \cos(t)\), \(y= 2\sin(t)\). Eliminate \(t\) to find the Cartesian equation.
Use \(x^2 + \bigl(\frac{y}{2}\bigr)^2= \cos^2(t)+ \sin^2(t)=1.\)
\(\frac{x^2}{1}+ \frac{y^2}{4}=1\). It's an ellipse.
Question 75
Prove the identity: \(\sin(x)+ \sin(y)= 2\,\sin\Bigl(\frac{x+y}{2}\Bigr)\cos\Bigl(\frac{x-y}{2}\Bigr).\)
This is a standard sum-to-product identity for sine.
Suppose a triangle with sides \(a,b,c\) has angles \(A,B,C\). Prove that \(a : b : c= \sin(A) : \sin(B) : \sin(C).\)
Use the Law of Sines: \(\frac{a}{\sin(A)}= \frac{b}{\sin(B)}= \frac{c}{\sin(C)}= 2R\) (the circumradius).
By the Law of Sines, \(\frac{a}{\sin(A)}= \frac{b}{\sin(B)}= \frac{c}{\sin(C)}\implies a : b : c= \sin(A) : \sin(B) : \sin(C).\)
Question 77
Prove the identity for hyperbolic sine addition: \(\sinh(x+y)= \sinh(x)\cosh(y)+ \cosh(x)\sinh(y).\)
Use definitions: \(\sinh(x)= \frac{e^x- e^{-x}}{2}, \cosh(x)= \frac{e^x+ e^{-x}}{2}\).
Follows from \(\sinh(x+y)= \frac{e^{x+y}- e^{-(x+y)}}{2}= \frac{e^x e^y - e^{-x} e^{-y}}{2}\). Distribute and regroup to get \(\sinh(x)\cosh(y)+ \cosh(x)\sinh(y).\)
Question 78
Prove the identity for hyperbolic cosine addition: \(\cosh(x+y)= \cosh(x)\cosh(y)+ \sinh(x)\sinh(y).\)
Similar approach to the previous: use expansions of \(e^x, e^y\).