Question

Precalc - trig identities. Decent explanation, please. THX!!!

Answer 1

12. Recall the half-angle identity,

$\sin^2\dfrac y2=\dfrac{1-\cos y}2\implies\sin\dfrac y2=\sqrt{\dfrac{1-\cos y}2}$

where we take the positive square root because $y$ is an angle in a right triangle, which means $0^\circ$, so $0^\circ$. For such an angle, it's always the case that $\sin\dfrac y2>0$.

Use the Pythagorean theorem to find the length of the hypotenuse:

$\sqrt{5^2+12^2}=\sqrt{169}=13$

Then

$\sin\dfrac y2=\sqrt{\dfrac{1-\frac5{13}}2}=\sqrt{\dfrac4{13}}=\dfrac2{\sqrt{13}}$

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13. $x$ is in quadrant II, which means $90^\circ$, so $45^\circ$. In other words, $\dfrac x2$ is in quadrant I, so $\sin\dfrac x2>0$. From the half-angle identity we get

$\sin\dfrac x2=\sqrt{\dfrac{1-\cos x}2}=\sqrt{\dfrac23}$

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14. Simplification follows from the definitions of each function:

$\sec x=\dfrac1{\cos x}$

$\tan x=\dfrac{\sin x}{\cos x}$

$\csc x=\dfrac1{\sin x}$

So we have

$\sec x\tan x\cos x\csc x=\dfrac{\sin x\cos x}{\cos^2x\sin x}=\dfrac1{\cos x}$

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15. Use the Pythagorean identity:

$\cos^2x+\sin^2x=1\implies\dfrac{\cos^2x}{\cos^2x}+\dfrac{\sin^2x}{\cos^2x}=\dfrac1{\cos^2x}\implies1+\tan^2x=\sec^2x$

Then

$\tan^3x+\tan x=\tan x(\tan^2x+1)=\tan x\sec^2x$