Question

sin 2(theta) + cos 2(theta) gives what

Answer 1

Are they asking you to simplify using the double-angle formula?

Answer 2

they are asking to complete the formula
just complete the formula

Answer 3

= 1 (if they are in square form)

Answer 4

@waterineyes
it is double ang

Answer 5

\[\huge \sin^ \theta + \cos^2 \theta = 1\]

Answer 6

Oh wait, is it \[\sin^2(\theta)+\cos^2(\theta)\] or
\[\sin(2\theta)+\cos(2\theta)\]

Answer 7

second one @mathteacher1729

Answer 8

Gotcha. Then it's the double-angle formula. All you have to do is find the double-angle formula (I recommend this cheat sheet http://tutorial.math.lamar.edu/cheat_table.aspx ) and then replace each term (sin and cos) with the appropriate expression.

Answer 9

\[= 2\sin \theta \cos \theta + \cos^2 \theta - \sin^2 \theta\]

Answer 10

\(\sin2A = 2 \sin A\cos A\)
\(\cos2A = \cos^2A - \sin^2A\)

so, \(sin2A + cos 2A = 2 \sin A\cos A + \cos^2A - \sin^2A\)
or, \(sin2A + cos 2A = (2 \sin A\cos A + \cos^2A + \sin^2A) - 2 \sin^2A\)
or, \(sin2A + cos 2A = (\cos A + \sin A)^2 - 2 \sin^2A\)

Answer 11

I can't seem to think how this can be simplified further. Or may be the first step itself is the most simplified form possible.

Answer 12

Moreover,

\[\large = (cosA + sinA - \sqrt{2}sinA)(cosA + sinA + \sqrt{2}\sin A)\]

Answer 13

yeah, in factorised format^

Answer 14

\[2\tan \theta + (1-\tan ^{2}\theta) \div (1+\tan ^{2}\theta )\]

Answer 15

this is what she wanted

Answer 16

She???

Answer 17

miss

Answer 18

Ok I got it now...

Answer 19

i will give u one ques which u cant do at all

Answer 20

Use these identities:

\[\huge \color{green}{\sin2 \theta = \frac{2\tan \theta}{1 + \tan^2 \theta}}\]

\[\huge \color{green}{\cos2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}}\]

Answer 21

Just take the LCM:

\[\large = \frac{2 \tan \theta + (1 - \tan^2 \theta)}{1 + \tan^2 \theta}\]

Got it??

Answer 22

i got it before u man

Answer 23

Oh that is so nice then..