Math  /  Trigonometry

Question1) Simplify. a) sinx(1cosx)\sin x\left(\frac{1}{\cos x}\right) b) (cosx)(secx)(\cos x)(\sec x) c) 1cos2x1-\cos ^{2} x d) 1sin2x1-\sin ^{2} x e) tanxsinx\frac{\tan x}{\sin x} f) (1sinx)(1+sinx)(1-\sin x)(1+\sin x) g) (1tanx)sinx\left(\frac{1}{\tan x}\right) \sin x h) 1+tan2xtan2x\frac{1+\tan ^{2} x}{\tan ^{2} x} i) sinxcosx1sin2x\frac{\sin x \cos x}{1-\sin ^{2} x} j) 1cos2xsinxcosx\frac{1-\cos ^{2} x}{\sin x \cos x}

Studdy Solution

STEP 1

What is this asking? We're going to simplify a bunch of trigonometric expressions using trig identities! Watch out! Remember your fundamental trig identities and be careful with your algebra!

STEP 2

1. Simplify a
2. Simplify b
3. Simplify c
4. Simplify d
5. Simplify e
6. Simplify f
7. Simplify g
8. Simplify h
9. Simplify i
10. Simplify j

STEP 3

We're given sinx1cosx\sin x \cdot \frac{1}{\cos x}.
Since 1cosx\frac{1}{\cos x} is the same as secx\sec x, we can rewrite this as sinxsecx\sin x \cdot \sec x.
We also know that secx\sec x is 1cosx\frac{1}{\cos x}, so we have sinx1cosx=sinxcosx\sin x \cdot \frac{1}{\cos x} = \frac{\sin x}{\cos x}.
And guess what? sinxcosx\frac{\sin x}{\cos x} is the definition of tanx\tan x!

STEP 4

We have (cosx)(secx)(\cos x)(\sec x).
Remember that secx=1cosx\sec x = \frac{1}{\cos x}.
So, we can rewrite our expression as (cosx)1cosx(\cos x) \cdot \frac{1}{\cos x}.
This simplifies to cosxcosx\frac{\cos x}{\cos x}, which is just **1**!

STEP 5

We're given 1cos2x1 - \cos^2 x.
One of the most important trig identities is sin2x+cos2x=1\sin^2 x + \cos^2 x = 1.
If we subtract cos2x\cos^2 x from both sides of this identity, we get 1cos2x=sin2x1 - \cos^2 x = \sin^2 x.
So, our expression simplifies to sin2x\sin^2 x!

STEP 6

We have 1sin2x1 - \sin^2 x.
Using the Pythagorean identity sin2x+cos2x=1\sin^2 x + \cos^2 x = 1, we can subtract sin2x\sin^2 x from both sides to get cos2x=1sin2x\cos^2 x = 1 - \sin^2 x.
So, our expression simplifies to cos2x\cos^2 x!

STEP 7

We're given tanxsinx\frac{\tan x}{\sin x}.
Remember that tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}.
So, we can rewrite our expression as sinxcosxsinx\frac{\frac{\sin x}{\cos x}}{\sin x}, which is the same as sinxcosx1sinx\frac{\sin x}{\cos x} \cdot \frac{1}{\sin x}.
The sinx\sin x terms divide to one, leaving us with 1cosx\frac{1}{\cos x}, which is secx\sec x!

STEP 8

We have (1sinx)(1+sinx)(1 - \sin x)(1 + \sin x).
This looks like a difference of squares!
Multiplying it out gives us 1sin2x1 - \sin^2 x.
And we know from before that 1sin2x=cos2x1 - \sin^2 x = \cos^2 x!

STEP 9

We have (1tanx)sinx\left(\frac{1}{\tan x}\right) \sin x.
Since 1tanx\frac{1}{\tan x} is cotx\cot x, we have cotxsinx\cot x \cdot \sin x.
We also know that cotx=cosxsinx\cot x = \frac{\cos x}{\sin x}.
So, our expression becomes cosxsinxsinx\frac{\cos x}{\sin x} \cdot \sin x.
The sinx\sin x terms divide to one, leaving us with cosx\cos x!

STEP 10

We're given 1+tan2xtan2x\frac{1 + \tan^2 x}{\tan^2 x}.
We can rewrite this as 1tan2x+tan2xtan2x\frac{1}{\tan^2 x} + \frac{\tan^2 x}{\tan^2 x}.
This simplifies to 1tan2x+1\frac{1}{\tan^2 x} + 1.
Since 1tanx=cotx\frac{1}{\tan x} = \cot x, we have cot2x+1\cot^2 x + 1.
And remember the identity 1+cot2x=csc2x1 + \cot^2 x = \csc^2 x?
So, our expression simplifies to csc2x\csc^2 x!

STEP 11

We have sinxcosx1sin2x\frac{\sin x \cos x}{1 - \sin^2 x}.
We know that 1sin2x=cos2x1 - \sin^2 x = \cos^2 x, so we can rewrite our expression as sinxcosxcos2x\frac{\sin x \cos x}{\cos^2 x}.
This simplifies to sinxcosx\frac{\sin x}{\cos x} by dividing the numerator and denominator by cosx\cos x, which is just tanx\tan x!

STEP 12

We're given 1cos2xsinxcosx\frac{1 - \cos^2 x}{\sin x \cos x}.
We know 1cos2x=sin2x1 - \cos^2 x = \sin^2 x, so we can rewrite our expression as sin2xsinxcosx\frac{\sin^2 x}{\sin x \cos x}.
This simplifies to sinxcosx\frac{\sin x}{\cos x} by dividing the numerator and denominator by sinx\sin x, which is tanx\tan x!

STEP 13

a) tanx\tan x b) 1 c) sin2x\sin^2 x d) cos2x\cos^2 x e) secx\sec x f) cos2x\cos^2 x g) cosx\cos x h) csc2x\csc^2 x i) tanx\tan x j) tanx\tan x

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