Question

cos^2A -sin^2A=1/2 then cos^4A-sin^4A=?

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Answer 1

Step-by-step explanation:

It's in the attachment :)

Answer 2

Given,

$\longrightarrow\cos^2A-\sin^2A=\dfrac{1}{2}\quad\quad\dots(1)$

We have to find value of $\cos^4A-\sin^4A.$

We know that,

$\longrightarrow\cos^2A+\sin^2A=1\quad\quad\dots(2)$

So, on multiplying (1) and (2),

$\longrightarrow(\cos^2A-\sin^2A)(\cos^2A+\sin^2A)=\dfrac{1}{2}\times1$

Since $(a-b)(a+b)=a^2-b^2,$

$\longrightarrow(\cos^2A)^2-(\sin^2A)^2=\dfrac{1}{2}$

$\longrightarrow\underline{\underline{\cos^4A-\sin^4A=\dfrac{1}{2}}}$

Hence 1/2 is the answer.

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Some Trigonometric Identities:-

$\longrightarrow\sin^2A+\cos^2A=1$

$\longrightarrow\sec^2A-\tan^2A=1$

$\longrightarrow\csc^2A-\cot^2A=1$

$\longrightarrow\sin(A\pm B)=\sin A\cos B\pm \cos A\sin B$

$\longrightarrow\cos(A\pm B)=\cos A\cos B\mp \sin A\sin B$

$\longrightarrow\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

$\longrightarrow \sin(A+B)+\sin(A-B)=2\sin A\cos B$

$\longrightarrow \sin(A+B)-\sin(A-B)=2\cos A\sin B$

$\longrightarrow \cos(A+B)+\cos(A-B)=2\cos A\cos B$

$\longrightarrow \cos(A+B)-\cos(A-B)=-2\sin A\sin B$