Is 5cos^2(x)+7sin^2(x)-5=2sin^2(x) ?

The answer to whether 5cos^2(x)+7sin^2(x)-5=2sin^2(x) is True

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prove 5cos^2(x)+7sin^2(x)-5=2sin^2(x)

prove

5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

True

Solution steps

5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

Manipulating left side

5 cos^{2} (x) + 7 sin^{2} (x) - 5

Rewrite using trig identities

5 cos^{2} (x) + 7 sin^{2} (x) - 5

Use the Pythagorean identity:

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

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

= 5 (1 - sin^{2} (x)) + 7 sin^{2} (x) - 5

Simplify

5 (1 - sin^{2} (x)) + 7 sin^{2} (x) - 5 : 2 sin^{2} (x)

5 (1 - sin^{2} (x)) + 7 sin^{2} (x) - 5

Expand

5 (1 - sin^{2} (x)) : 5 - 5 sin^{2} (x)

5 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 1, c = sin^{2} (x)

= 5 \cdot 1 - 5 sin^{2} (x)

Multiply the numbers:

5 \cdot 1 = 5

= 5 - 5 sin^{2} (x)

= 5 - 5 sin^{2} (x) + 7 sin^{2} (x) - 5

Simplify

5 - 5 sin^{2} (x) + 7 sin^{2} (x) - 5 : 2 sin^{2} (x)

5 - 5 sin^{2} (x) + 7 sin^{2} (x) - 5

Add similar elements:

- 5 sin^{2} (x) + 7 sin^{2} (x) = 2 sin^{2} (x)

= 5 + 2 sin^{2} (x) - 5

Group like terms

= 2 sin^{2} (x) + 5 - 5

5 - 5 = 0

= 2 sin^{2} (x)

= 2 sin^{2} (x)

= 2 sin^{2} (x)

= 2 sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (θ) - sin (θ - 2 π) = 0

- sin (2 x) = - 2 sin (x) cos (x)

csc (2 x) = \frac{sec ( x ) csc ( x )}{2}

sec (2 x) = \frac{( sec ^{2} ( x ) )}{2 - sec ^{2} ( x )}

sin (x) = 0

Is 5cos^2(x)+7sin^2(x)-5=2sin^2(x) ?
The answer to whether 5cos^2(x)+7sin^2(x)-5=2sin^2(x) is True