"Trigonometry" — Practice

Question 1

Find the exact value of $\cos 75^\circ$.

Accepted Answer

Write $75^\circ$ as $45^\circ + 30^\circ$ and use the compound-angle formula, remembering that cosine of a sum takes a minus sign: $\cos45\cos30 - \sin45\sin30 = (\sqrt2/2)(\sqrt3/2) - (\sqrt2/2)(1/2) = (\sqrt6-\sqrt2)/4$, about 0.2588. As a check, this equals $\sin15^\circ$, which it must, since $75$ and $15$ are complementary.

Question 2

Prove that $\cos 3	heta = 4\cos^{3}	heta - 3\cos	heta$.

Accepted Answer

Split $3	heta$ as $2	heta + 	heta$ and expand: $\cos3	heta = \cos2	heta\cos	heta - \sin2	heta\sin	heta$. Substituting $\cos2	heta = 2\cos^2	heta - 1$ and $\sin2	heta = 2\sin	heta\cos	heta$ gives $(2\cos^2	heta-1)\cos	heta - 2\sin^2	heta\cos	heta$. Replace $\sin^2	heta$ by $1-\cos^2	heta$ so only cosines remain: $2\cos^3	heta - \cos	heta - 2\cos	heta + 2\cos^3	heta = 4\cos^3	heta - 3\cos	heta$.

Question 3

If $A+B+C=\pi$, prove that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$.

Accepted Answer

Group the first two and convert the sum to a product: $\sin2A+\sin2B = 2\sin(A+B)\cos(A-B)$. Since $A+B = \pi - C$, this is $2\sin C\cos(A-B)$. The third term is $\sin2C = 2\sin C\cos C$, and $\cos C = -\cos(A+B)$. So the total is $2\sin C[\cos(A-B) - \cos(A+B)]$, and the bracket is $2\sin A\sin B$ by the difference-to-product formula. The whole is $4\sin A\sin B\sin C$.

Question 4

Find the maximum and minimum values of $5\sin	heta - 12\cos	heta$.

Accepted Answer

The amplitude is the square root of $25 + 144 = 169$, which is 13, so the expression ranges over the closed interval from $-13$ to $13$. The maximum is 13 and the minimum is $-13$. Note that $5+12 = 17$ is not attainable, because the sine and cosine terms reach their extremes at different angles and cannot both peak together.

Question 5

Solve $2\sin^{2}	heta - 3\cos	heta = 0$.

Accepted Answer

Convert to a single ratio using the Pythagorean identity: $2(1-\cos^2	heta) - 3\cos	heta = 0$, that is $2\cos^2	heta + 3\cos	heta - 2 = 0$, which factorises as $(2\cos	heta - 1)(\cos	heta + 2) = 0$. The second factor gives $\cos	heta = -2$, which is impossible and must be rejected explicitly. The first gives $\cos	heta = 1/2$, so $	heta = 2n\pi \pm \pi/3$.

Question 6

Solve $\sin	heta + \sin 3	heta + \sin 5	heta = 0$.

Accepted Answer

Group the outer two, whose half-sum is $3	heta$: $\sin	heta + \sin5	heta = 2\sin3	heta\cos2	heta$. The equation becomes $\sin3	heta(2\cos2	heta + 1) = 0$. From the first factor, $3	heta = n\pi$, so $	heta = n\pi/3$. From the second, $\cos2	heta = -1/2$, so $2	heta = 2n\pi \pm 2\pi/3$ and $	heta = n\pi \pm \pi/3$. Both families are required. Choosing which pair to group is the whole skill: the outer terms were chosen because their half-sum reproduces the middle angle.

Question 7

Solve $	an	heta + 	an 2	heta + 	an	heta	an2	heta = 1$.

Accepted Answer

Rearrange as $	an	heta + 	an2	heta = 1 - 	an	heta	an2	heta$, then divide both sides by the right-hand side to produce the tangent addition formula: $	an(	heta + 2	heta) = 1$, that is $	an3	heta = 1$. Hence $3	heta = n\pi + \pi/4$ and $	heta = n\pi/3 + \pi/12$. Recognising the addition formula hiding in the arrangement is what keeps this to two lines; expanding $	an2	heta$ instead produces a cubic.

Question 8

Evaluate $\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$.

Accepted Answer

The range of $\cos^{-1}$ is $[0,\pi]$, and $7\pi/6$ exceeds $\pi$, so the answer cannot be $7\pi/6$. Use $\cos(2\pi - x) = \cos x$: $\cos(7\pi/6) = \cos(2\pi - 7\pi/6) = \cos(5\pi/6)$, and $5\pi/6$ does lie in $[0,\pi]$. So the value is $5\pi/6$. Checking numerically, both angles have cosine $-\sqrt3/2$, and only one of them is in range.

Question 9

Evaluate $	an^{-1}\left(	an\dfrac{5\pi}{6}ight)$ and $\sin^{-1}\left(\sin\dfrac{5\pi}{6}ight)$.

Accepted Answer

For the tangent, the range is the open interval from $-\pi/2$ to $\pi/2$, and $5\pi/6$ is outside it. Subtract the period $\pi$: $5\pi/6 - \pi = -\pi/6$, which is in range, so the answer is $-\pi/6$. For the sine, the range is $[-\pi/2,\pi/2]$ and $5\pi/6$ is again outside. Use $\sin(\pi - x) = \sin x$: $\pi - 5\pi/6 = \pi/6$, so the answer is $\pi/6$. The same input gives two different adjustments because the two ranges differ.

Question 10

Prove that $	an^{-1}\dfrac{1}{3} + 	an^{-1}\dfrac{1}{5} + 	an^{-1}\dfrac{1}{7} + 	an^{-1}\dfrac{1}{8} = \dfrac{\pi}{4}$.

Accepted Answer

Combine in pairs, checking the condition each time. First, $xy = 1/15 < 1$, so $	an^{-1}(1/3)+	an^{-1}(1/5) = 	an^{-1}[(1/3+1/5)/(1-1/15)] = 	an^{-1}(8/15 \div 14/15) = 	an^{-1}(4/7)$. Next, $	an^{-1}(1/7)+	an^{-1}(1/8)$ has $xy = 1/56 < 1$ and gives $	an^{-1}(15/55) = 	an^{-1}(3/11)$. Finally $(4/7)(3/11) = 12/77 < 1$, so the sum is $	an^{-1}[(4/7+3/11)/(1-12/77)] = 	an^{-1}(65/77 \div 65/77) = 	an^{-1}1 = \pi/4$.

Question 11

If $\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \dfrac{3\pi}{2}$, find $x+y+z$.

Accepted Answer

Each inverse sine is at most $\pi/2$, so three of them can total $3\pi/2$ only if every one is exactly $\pi/2$. That forces $x = y = z = 1$, so $x+y+z = 3$. This is an extremal argument rather than an algebraic one: the constraint saturates the maximum, leaving no freedom at all. Attempting to expand with addition formulas here wastes several minutes and reaches the same place.

Question 12

Solve $\cos	heta - \sin	heta = 1$, and identify any roots created by squaring.

Accepted Answer

Squaring gives $1 - 2\sin	heta\cos	heta = 1$, so $\sin2	heta = 0$ and $	heta = n\pi/2$. Within one revolution the candidates are $0$, $\pi/2$, $\pi$ and $3\pi/2$. Substituting into the original: $	heta = 0$ gives $1-0 = 1$, valid; $	heta = \pi/2$ gives $0-1 = -1$, false; $	heta = \pi$ gives $-1-0 = -1$, false; $	heta = 3\pi/2$ gives $0+1 = 1$, valid. So the solutions are $	heta = 2n\pi$ and $	heta = 2n\pi - \pi/2$, and the other two were manufactured by the squaring step.

Practice — "Trigonometry"

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