Precalc trig identities! cos 2x + sin 2x = 1?

Question

Solve the following equation:

cos 2x + sin 2x = 1

Answer 1

There are double angle formulas that you can use to substitute cos2x and sin2x.

cos2x=cos^2x-sin^2x
sin2x=2sinxcosx

So,
cos2x+sin2x=cos^2x-sin^2x+2sinxcosx=1

Now, we know that cos^2x=1-sin^2x

1-sin^2x-sin^2x+2sinxcosx-1=0
-2sin^2x+2sinxcosx=0
Factor out a -2sinx

-2sinx(sinx-cosx)=0

So, when does sinx=0? When does sinx=cosx.

sinx=0

x=0 + pi (n) <--n=any integer.

sinx=cosx

x=pi/4+pi(n)

Altogether,

x=0+pi(n), (pi/4) +pi(n)

I hope that this helps.

Answer 2

Note: There is an identity cos^2 x + sin^2 x = 1. Since it is an identity, it means that it is true for every x.

If you actually mean cos (2x) + sin (2x) = 1, then write sin (2x) = 1 - cos(2x) and substitute into cos^2 (2x) + sin^2 (2x) = 1 [from the identity]. Then you get
cos^2 (2x) + (1 - cos (2x))^2 = 1
=> cos^2 (2x) + 1 - 2 cos (2x) + cos^2 (2x) = 1
=> 2 cos^2 (2x) - 2 cos (2x) = 0
=> 2 cos (2x) [cos (2x) - 1] = 0
=> cos (2x) = 0 or cos (2x) = 1
So either cos(2x) = 1 and sin(2x) = 0, or sin(2x) = 1 and cos(2x) = 0. This can be simplified to saying that either cos(2x) = 1 or sin(2x) = 1.

cos 2x = 1 => 2x = 2kπ for some integer k
sin 2x = 1 => 2x = (2n + 1/2)π for some integer n.

Hence solutions for x are {kπ, k ∈ Z} ∪ {(n + 1/4)π, n ∈ Z}.

Answer 3

This page shows the derivations of the three Pythagorean Identities. A "derivation" means that we need to create this from scratch - or, at least, from other things that we know.

http://www.coolmath.com/pythagoreanidentities.htm

Proof of the Pythagorean identities

To prove:

a) sin²θ + cos²θ = 1

b) 1 + tan²θ = sec²θ

c) 1 + cot²θ = csc ²θ

Proof. According to the Pythagorean theorem,

x² + y² = r². . . . . . . . . . . . . . . .(1)

Therefore, on dividing both sides by r²,

x²/r² + y²/r² = r²/r² = 1.

That is, according to the definitions,

cos²θ + sin²θ = 1.

Apart from the order of the terms, this is the first Pythagorean identity, a). To derive b), divide line (1) by x²; and to derive c), divide by y².

Answer 4

sin(?+?) = sin(?)cos(?) + cos(?)sin(?), so sin(2x)cos(fifty 5) + cos(2x)sin(fifty 5) = sin(2x+fifty 5). sin(2x+fifty 5) = a million/2, so 2x + fifty 5 = sin?¹(a million/2) = 60°. 2x + fifty 5 = 60°, or 2x = 5°, or x = 2.5°.