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Math  /  Trigonometry

Question1−2sin⁡2xsin⁡x+cos⁡x+2sin⁡x2cos⁡x2=cos⁡x\frac{1-2 \sin ^{2} x}{\sin x+\cos x}+2 \sin \frac{x}{2} \cos \frac{x}{2}=\cos x

Studdy Solution

STEP 1

What is this asking? Can we show that this crazy-looking fraction with sines and cosines, plus another term with sines and cosines of half angles, simplifies to just plain cosine of xx? Watch out! Remember those trig identities, especially the double-angle and half-angle formulas!
Also, be super careful with signs when simplifying.

STEP 2

1. Simplify the First Term
2. Simplify the Second Term
3. Combine and Conquer

STEP 3

Alright, let's **tackle that first term**, the fraction!
Notice that the numerator 1−2sin⁡2x1 - 2 \sin^2 x looks awfully familiar.
It's the **double-angle identity for cosine**: cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2 \sin^2 x.
So, let's **swap that in**!

STEP 4

Now our fraction is cos⁡2xsin⁡x+cos⁡x\frac{\cos 2x}{\sin x + \cos x}.
But cos⁡2x\cos 2x also has another identity: cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x.
This is a **difference of squares**, which we can factor!
Let's **rewrite** the fraction as (cos⁡x−sin⁡x)(cos⁡x+sin⁡x)sin⁡x+cos⁡x\frac{(\cos x - \sin x)(\cos x + \sin x)}{\sin x + \cos x}.

STEP 5

Whoa, look at that!
We have cos⁡x+sin⁡x\cos x + \sin x on both the top and bottom.
Since sin⁡x+cos⁡x\sin x + \cos x is the same as cos⁡x+sin⁡x\cos x + \sin x, we can **divide to one**!
This leaves us with just cos⁡x−sin⁡x\cos x - \sin x.
Nice!

STEP 6

Now for the **second term**: 2sin⁡x2cos⁡x22 \sin \frac{x}{2} \cos \frac{x}{2}.
This is screaming **double-angle identity**, but for sine!
Remember, sin⁡2a=2sin⁡acos⁡a\sin 2a = 2 \sin a \cos a.
Here, our "a" is x2\frac{x}{2}, so 2sin⁡x2cos⁡x2=sin⁡(2⋅x2)=sin⁡x2 \sin \frac{x}{2} \cos \frac{x}{2} = \sin \left(2 \cdot \frac{x}{2}\right) = \sin x.
Boom!

STEP 7

Let's **put it all together**!
We simplified the first term to cos⁡x−sin⁡x\cos x - \sin x and the second term to sin⁡x\sin x.
So the whole left side of the equation is now (cos⁡x−sin⁡x)+sin⁡x(\cos x - \sin x) + \sin x.

STEP 8

We can **add −sin⁡x-\sin x and sin⁡x\sin x to zero**, which leaves us with just cos⁡x\cos x.
And guess what?
That's exactly the right side of the equation!
We did it!

STEP 9

We've shown that 1−2sin⁡2xsin⁡x+cos⁡x+2sin⁡x2cos⁡x2\frac{1-2 \sin ^{2} x}{\sin x+\cos x}+2 \sin \frac{x}{2} \cos \frac{x}{2} simplifies to cos⁡x\cos x, which is what we wanted to prove.

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