Is 7cos^2(x)+5sin^2(x)=cos(2x)+6 ?

The answer to whether 7cos^2(x)+5sin^2(x)=cos(2x)+6 is True

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prove 7cos^2(x)+5sin^2(x)=cos(2x)+6

prove

7 cos^{2} (x) + 5 sin^{2} (x) = cos (2 x) + 6

True

Solution steps

7 cos^{2} (x) + 5 sin^{2} (x) = cos (2 x) + 6

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 7, b = 1, c = sin^{2} (x)

= 7 \cdot 1 - 7 sin^{2} (x)

Multiply the numbers:

7 \cdot 1 = 7

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

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

Add similar elements:

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

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

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

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

Manipulating right side

cos (2 x) + 6

Rewrite using trig identities

cos (2 x) + 6

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (x) + 6

Simplify

1 - 2 sin^{2} (x) + 6 : - 2 sin^{2} (x) + 7

1 - 2 sin^{2} (x) + 6

Group like terms

= - 2 sin^{2} (x) + 1 + 6

Add the numbers:

1 + 6 = 7

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

1 + sec^{2} (x) = 2 + tan^{2} (x)

9 (1 + sin (y)) (1 + sin (- y)) = 9 cos^{2} (y)

(\frac{tan ( θ )}{sec ( θ )}) = sin (θ)

sin (4 5^{\circ}) cot (4 5^{\circ}) = cos (4 5^{\circ})

sec (4 x) = 2

Is 7cos^2(x)+5sin^2(x)=cos(2x)+6 ?
The answer to whether 7cos^2(x)+5sin^2(x)=cos(2x)+6 is True