Is 7cos^2(b)-7sin^2(b)=14cos^2(b)-7 ?

The answer to whether 7cos^2(b)-7sin^2(b)=14cos^2(b)-7 is True

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prove 7cos^2(b)-7sin^2(b)=14cos^2(b)-7

prove

7 cos^{2} (b) - 7 sin^{2} (b) = 14 cos^{2} (b) - 7

True

Solution steps

7 cos^{2} (b) - 7 sin^{2} (b) = 14 cos^{2} (b) - 7

Manipulating left side

7 cos^{2} (b) - 7 sin^{2} (b)

Rewrite using trig identities

7 cos^{2} (b) - 7 sin^{2} (b)

Use the Pythagorean identity:

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

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

= 7 cos^{2} (b) - 7 (1 - cos^{2} (b))

Simplify

7 cos^{2} (b) - 7 (1 - cos^{2} (b)) : 14 cos^{2} (b) - 7

7 cos^{2} (b) - 7 (1 - cos^{2} (b))

Expand

- 7 (1 - cos^{2} (b)) : - 7 + 7 cos^{2} (b)

- 7 (1 - cos^{2} (b))

Apply the distributive law:

a (b - c) = ab - a c

a = - 7, b = 1, c = cos^{2} (b)

= - 7 \cdot 1 - (- 7) cos^{2} (b)

Apply minus-plus rules

- (- a) = a

= - 7 \cdot 1 + 7 cos^{2} (b)

Multiply the numbers:

7 \cdot 1 = 7

= - 7 + 7 cos^{2} (b)

= 7 cos^{2} (b) - 7 + 7 cos^{2} (b)

Simplify

7 cos^{2} (b) - 7 + 7 cos^{2} (b) : 14 cos^{2} (b) - 7

7 cos^{2} (b) - 7 + 7 cos^{2} (b)

Group like terms

= 7 cos^{2} (b) + 7 cos^{2} (b) - 7

Add similar elements:

7 cos^{2} (b) + 7 cos^{2} (b) = 14 cos^{2} (b)

= 14 cos^{2} (b) - 7

= 14 cos^{2} (b) - 7

= 14 cos^{2} (b) - 7

= 14 cos^{2} (b) - 7

We showed that the two sides could take the same form

\Rightarrow True

1 + tan (x) \cdot tan (2 x) = sec (2 x)

0 sec (0) = tan (0)

tan^{3} (x) (csc^{3} (x)) = sec^{3} (x)

sin (x) sec (x) = sin (x)

tan^{2} (y) - sin^{2} (y) = tan^{2} (y) sin^{2} (y)

Is 7cos^2(b)-7sin^2(b)=14cos^2(b)-7 ?
The answer to whether 7cos^2(b)-7sin^2(b)=14cos^2(b)-7 is True