Question

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

Answer 1

12. Recall the half-angle identity,

`\sin^2\frac y2=\frac{1-\cos y}2\implies\sin\frac y2=\sqrt{\frac{1-\cos y}2}`

where we take the positive square root because
`y` is an angle in a right triangle, which means
`0^\circ<y<90^\circ`, so
`0^\circ<\frac y2<45^\circ`. For such an angle, it's always the case that
`\sin\frac y2>0`.

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

`√(5^2+12^2)=√(169)=13`

Then

`\sin\frac y2=\sqrt{\frac{1-\frac5{13}}2}=\sqrt{\frac4{13}}=\frac2{√(13)}`

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13.
`x` is in quadrant II, which means
`90^\circ<x<180^\circ`, so
`45^\circ<\frac x2<90^\circ`. In other words,
`\frac x2` is in quadrant I, so
`\sin\frac x2>0`. From the half-angle identity we get

`\sin\frac x2=\sqrt{\frac{1-\cos x}2}=√(\frac23)`

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14. Simplification follows from the definitions of each function:

`\sec x=\frac1{\cos x}`

`\tan x=(\sin x)/(\cos x)`

`\csc x=\frac1{\sin x}`

So we have

`\sec x\tan x\cos x\csc x=(\sin x\cos x)/(\cos^2x\sin x)=\frac1{\cos x}`

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15. Use the Pythagorean identity:

`\cos^2x+\sin^2x=1\implies(\cos^2x)/(\cos^2x)+(\sin^2x)/(\cos^2x)=\frac1{\cos^2x}\implies1+\tan^2x=\sec^2x`

Then

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