Double The Angle

We know that $\cos2\theta = \cos^2\theta - \sin^2\theta.$

Since $\cos\theta = \sin \theta,$ $\cos^2\theta = \sin^2 \theta.$

Thus, $\cos^2\theta - \sin^2 \theta = \cos2 \theta = 0.$

I don't know if it's correct, i did it like this $sin \theta = cos \theta$

Divide by $cos \theta$ , we'll get $tan \theta = 1$

Theta value for tan is 1 only when it is $45 ^ \circ$

Taking $2 \times 45^ \circ = 90^ \circ$

$cos 90 ^ \circ = 0$

• There are two ways to solve this one.
• I : If $\sin \theta = \cos \theta$ , then $\theta$ can only equal 45º and 225º .
• $\cos (2 \times 45) = \cos 90 = 0$ or $\cos (2 \times 225) = \cos 450 = \cos (360 + 90) = 0$ .
• II : We know $\cos (a+b) = (\cos a \times \cos b) - (\sin a \times \sin b)$ .
• Using $\theta$ , we have $\cos 2\theta = (\cos \theta \times \cos \theta) - (\sin \theta \times \sin \theta)$ . $\cos 2\theta = \cos² \theta - \sin² \theta$ .
• Since $\cos \theta = \sin \theta$ , we have $\cos² \theta = \sin² \theta$ . $\cos 2\theta = \cos² \theta - \cos² \theta = 0$ .

when $\sin\theta = \cos\theta$ then the only possibility is $\sin 45^\circ =\cos 45^\circ$ because $\sin 45^\circ =\cos 45^\circ =\frac{1}{\sqrt{2}}$ therefore $\cos 2 \theta=\cos 90^\circ = 0$

Since $\cos 2 \theta = cos^2 \theta - \sin^2 \theta$ can be rewritten acording to $A^2 - B^2 = (A - B) (A + B)$ as $\cos 2 \theta = ( \cos \theta - \sin \theta ) ( \cos \theta + \sin \theta )$ and $\cos \theta = \sin \theta \Leftrightarrow \cos \theta - \sin \theta =0$ hence the expression $\cos 2 \theta = 0 (\cos \theta + \sin \theta) = 0$ .

Although not the best way, here is another way of working it out without numerical values.

Since $\cos { 2\theta } =2\cos { \theta } \sin { \theta }$

We get

$\cos { \theta } \sin { \theta } =2\cos { \theta } \sin { \theta }$

We divide by $\cos { \theta } \sin { \theta }$ , giving:

$2=0$

Obviously

$2\neq 0$

hence $\cos { \theta } \sin { \theta }$ is zero or undefined.

If $\cos { \theta } \sin { \theta }$ is zero then $\ 2 cos { \theta } \sin { \theta }$ is zero, giving $\cos { 2\theta } = 2 \cos { \theta } \sin { \theta } =0$

You could also say that a solution lies at $\theta =\infty$ but I feel like people would disagree.

We know when sinA=cosa is At A =45° cos2A=cos90° =0

I'm simple so I knew anything divided by zero is zero so cos2(0) is zero

Given that, sine theta = cos theta. This is possible when, theta= 45 degrees. Now, cos $2*theta= cos 2*45degree$ = cos 90degree = 0

I think this is the simplest way :D

Let $a,b,c$ be the sides of a rectangle triangle.

We know that $\sin \theta = \cos \theta$ is equivalent to $\frac{a}{c}=\frac{b}{c}$ and therefore $a=b$ , which means we are talking about a triangle with angles 90, 45 and 45.

Then, $\cos (2\theta) = \cos (2·45) =\boxed{0}$