Trigonometric proof

**Pandora** · August 11th 2008, 01:03 AM

I would like to prove that

sinX^4 - cosX^4 + 2cosX^2

equals to 1 using sin^2 theta+ cos^2 theta=1 and a^2-b^2=(a-b)(a+b).

I am completely stuck though, and do not know where to begin. Any advice would be appreciated.

**flyingsquirrel** · August 11th 2008, 01:09 AM

Hello

Note that $\sin^4x-\cos^4x=(\sin^2 x)^2-(\cos ^2 x)^2$ . Can you factor this using $a^2-b^2=(a-b)(a+b)$ ?

Hint (highlight to read) : * let a=(sin x)^2 and b=(cos x)^2 *

**Chris L T521** · August 11th 2008, 01:11 AM

Originally Posted by Pandora

I would like to prove that

sinX^4 - cosX^4 + 2cosX^2

equals to 1 using sin^2 theta+ cos^2 theta=1 and a^2-b^2=(a-b)(a+b).

I am completely stuck though, and do not know where to begin. Any advice would be appreciated.

$\sin^4(x)-\cos^4(x)+2\cos^2(x)=1$

Let's focus on this term first:

$\sin^4(x)-\cos^4(x)$

using the formula $a^2-b^2=(a+b)(a-b)$ , we see that

$\sin^4(x)-\cos^4(x)=(\sin^2(x)+\cos^2(x))(\sin^2(x)-\cos^2(x))=\sin^2(x)-\cos^2(x)$

Now, we see that the equation now becomes

$\sin^2(x)-\cos^2(x)+2\cos^2(x)=1$

Simplifying, we get

$\sin^2(x)+\cos^2(x)=1\implies 1=1$

Does this make sense?

--Chris

**Pandora** · August 11th 2008, 11:46 AM

Thank you both. It seems pretty clear now =) This forum is so efficient, and I will be definetly using it in the future.

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