0
$\begingroup$

My attempt:

$$1-2\sin^{2}x=\cos2x$$ Let $$\cos x-\sin x=t$$ Thus, on squaring: $$\sin2x=1-t^{2}$$

I tried simplifying further, but that $cos2x$ is giving me trouble.

Please help!

$\endgroup$
2
  • $\begingroup$ What is the original equation you are trying to solve? $\endgroup$ Mar 4 at 14:37
  • $\begingroup$ i mean it's the title itself $\endgroup$
    – Vega
    Mar 4 at 14:39
1
$\begingroup$

After a rewrite, we get $$(\cos x-\sin x)^3 = (\cos x-\sin x)(\cos x+\sin x)$$Can you solve from here?

$\endgroup$
5
  • $\begingroup$ I got two solutions, but i'm unable to get the third. $\endgroup$
    – Vega
    Mar 4 at 14:52
  • $\begingroup$ Which one can you not get? $\endgroup$ Mar 4 at 14:52
  • $\begingroup$ $x=2n\pi+\frac{\pi}{2}$ this one $\endgroup$
    – Vega
    Mar 4 at 14:53
  • $\begingroup$ $\sin\left(2n\pi+\frac\pi2\right)=\cos\left(2n\pi+\frac\pi2\right)$. What does that tell you in the above formula? $\endgroup$ Mar 4 at 14:55
  • $\begingroup$ Ok thanks, i found my mistake, all clear now! $\endgroup$
    – Vega
    Mar 4 at 15:04
0
$\begingroup$

$\sin2x=2\cos x\sin x \Rightarrow 1-\sin2x=cos^2x+sin^2x-2\sin x \cos x=(cosx-sinx)^2$

$1-\sin2x=(cosx-sinx)^2 \Rightarrow (1-\sin2x)(cosx-sinx)=(cosx-sinx)^3=1-2\sin^2x$

$(cosx-sinx)^3=1-2\sin^2x=\sin2x$

$cosx-sinx=t \Rightarrow t^3=1-t^2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.