0
$\begingroup$

how to proof that the following one is true

$\sin2 \theta + \cos2 \theta = \sin \theta + \cos \theta$

I tried to do like this

L.H.S.

$= 2\sin\theta\cos\theta + \cos^2 \theta - \sin^2 \theta $

$= \sin\theta\cos\theta + \cos^2 \theta + \sin\theta\cos\theta - \sin^2 \theta$

$= \cos \theta(\sin\theta + \cos \theta) + \sin \theta(\cos\theta - \sin \theta)$

Then what should I do ?

Am I on the right way ?

$\endgroup$
7
  • 10
    $\begingroup$ The statement is false,take $\theta =\frac { \pi }{ 4 } $,$$\\ \sin { \left( 2\cdot \frac { \pi }{ 4 } \right) } +\cos { \left( 2\cdot \frac { \pi }{ 4 } \right) } =\sin { \left( \frac { \pi }{ 4 } \right) } +\cos { \left( \frac { \pi }{ 4 } \right) } \\ 1\neq \sqrt { 2 } $$ $\endgroup$
    – haqnatural
    Commented Aug 29, 2017 at 13:12
  • $\begingroup$ I think this is not generally true . $\endgroup$
    – Iuli
    Commented Aug 29, 2017 at 13:13
  • 2
    $\begingroup$ This can not hold for all $\theta$ ! Take for example $\theta=\tfrac{\pi}{2}$. In fact you are asked to solve an equation : for which values of $\theta$ does the equality is valid... $\endgroup$
    – Jean Marie
    Commented Aug 29, 2017 at 13:14
  • $\begingroup$ oh really ? Thanks . I think in a wrong way $\endgroup$
    – why maths
    Commented Aug 29, 2017 at 13:16
  • $\begingroup$ Shouldn't the question be 'for which $\theta$ ... $\endgroup$
    – Pieter21
    Commented Aug 29, 2017 at 13:21

4 Answers 4

3
$\begingroup$

Let $(*)$ be the equation to solve.

$$ (*) \Leftrightarrow \sqrt2 \cos(2\theta-\frac{\pi}{4}) = \sqrt2 \cos(\theta-\frac{\pi}{4})$$

$$\Leftrightarrow 2\theta-\frac{\pi}{4} \equiv \theta-\frac{\pi}{4} \pmod{2\pi} \text{ or } 2\theta-\frac{\pi}{4} \equiv -\theta+\frac{\pi}{4} \pmod{2\pi}$$

$$ \Leftrightarrow \theta \equiv 0 \pmod{2\pi} \text{ or } \theta \equiv \frac{\pi}{6} \pmod{\frac{2\pi}{3}}$$

$\endgroup$
1
  • $\begingroup$ I think this was the intention. $\endgroup$
    – Pieter21
    Commented Aug 29, 2017 at 13:39
2
$\begingroup$

For $\theta=\pi$ you have

LHS: $\sin(2\pi) + \cos(2\pi) = 0 + 1 = 1$
RHS: $\sin\pi + \cos\pi = 0 + (-1) = -1$

hence the equality does not hold and can't be proven.

$\endgroup$
2
$\begingroup$

Use Simpson formula instead: $$ \sin(2x) + \cos(2x) = \sin(x) + \cos(x)\implies \sin(2x) - \sin(x) = \cos(x) - \cos(2x)\implies \\ \sin(x/2)( \cos(3x/2) - \sin(3x/2)) =0. $$

Then solve these two equations: $$ \sin(x/2) = 0 \\ \cos(3x/2) = \sin(3x/2). $$

$\endgroup$
2
  • $\begingroup$ Use LaTeX to render the math. $\endgroup$
    – Arash
    Commented Aug 29, 2017 at 14:46
  • $\begingroup$ Please strive not to vote to reopen dupes of FAQs, cf. recent site policy announcement here. $\endgroup$ Commented Jul 7, 2021 at 15:48
1
$\begingroup$

Shouldn't the question be 'for which $\theta$' ....

Then think along the unit circle.

Or use $\sin(x) + \cos(x) = \sqrt{2}\sin(x+\pi/4)$

Add $x$ and $y$ from any point, and compare it to the sum of $x'$ and $y'$ from the point at twice the angle.

Or use Wolfram Alpha to look for zeroes in: https://www.wolframalpha.com/input/?i=sin(x)+%2B+cos(x)+-+sin(2*x)+-+cos(2*x)

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .