Is 4cos^2(θ)+4sin^2(θ)=4 ?

The answer to whether 4cos^2(θ)+4sin^2(θ)=4 is True

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prove 4cos^2(θ)+4sin^2(θ)=4

prove

4 cos^{2} (θ) + 4 sin^{2} (θ) = 4

True

Solution steps

4 cos^{2} (θ) + 4 sin^{2} (θ) = 4

Manipulating left side

4 cos^{2} (θ) + 4 sin^{2} (θ)

Rewrite using trig identities

4 cos^{2} (θ) + 4 sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 4 cos^{2} (θ) + 4 (1 - cos^{2} (θ))

4 cos^{2} (θ) + 4 (1 - cos^{2} (θ)) = 4

4 cos^{2} (θ) + 4 (1 - cos^{2} (θ))

Expand

4 (1 - cos^{2} (θ)) : 4 - 4 cos^{2} (θ)

4 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= 4 cos^{2} (θ) + 4 - 4 cos^{2} (θ)

Simplify

4 cos^{2} (θ) + 4 - 4 cos^{2} (θ) : 4

4 cos^{2} (θ) + 4 - 4 cos^{2} (θ)

Group like terms

= 4 cos^{2} (θ) - 4 cos^{2} (θ) + 4

Add similar elements:

4 cos^{2} (θ) - 4 cos^{2} (θ) = 0

= 4

= 4

= 4

= 4

We showed that the two sides could take the same form

\Rightarrow True

\frac{- cos ^{2} ( 2 θ )}{2} = \frac{- 1 - cos ( 4 θ )}{8}

\frac{1 - cos ( x )}{- 1 + sec ( x )} = cos (x)

2 sin (y) cos (y) sec (2 y) = tan (2 y)

sec (x) + tan (x) = \frac{1}{sec ( x )} + tan (x)

tan (s) csc (s) = sec (s)