Is (sin(3a))/(sin(a))=2cos(2a)+1 ?

The answer to whether (sin(3a))/(sin(a))=2cos(2a)+1 is True

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prove (sin(3a))/(sin(a))=2cos(2a)+1

Solution

prove

\frac{sin ( 3 a )}{sin ( a )} = 2 cos (2 a) + 1

Solution

True

Solution steps

\frac{sin ( 3 a )}{sin ( a )} = 2 cos (2 a) + 1

Manipulating left side

\frac{sin ( 3 a )}{sin ( a )}

Rewrite using trig identities

\frac{sin ( 3 a )}{sin ( a )}

Use the following identity:

sin (3 x) = 3 sin (x) - 4 sin^{3} (x)

sin (3 x)

Rewrite using trig identities

sin (3 x)

Rewrite as

= sin (2 x + x)

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (2 x) cos (x) + cos (2 x) sin (x)

Use the Double Angle identity:

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

= cos (2 x) sin (x) + cos (x) 2 sin (x) cos (x)

Simplify

cos (2 x) sin (x) + cos (x) \cdot 2 sin (x) cos (x) : sin (x) cos (2 x) + 2 cos^{2} (x) sin (x)

cos (2 x) sin (x) + cos (x) 2 sin (x) cos (x)

cos (x) \cdot 2 sin (x) cos (x) = 2 cos^{2} (x) sin (x)

cos (x) 2 sin (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Expand

(1 - 2 sin^{2} (x)) sin (x) + 2 (1 - sin^{2} (x)) sin (x) : - 4 sin^{3} (x) + 3 sin (x)

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

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

Expand

sin (x) (1 - 2 sin^{2} (x)) : sin (x) - 2 sin^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

1 \cdot sin (x) - 2 sin^{2} (x) sin (x) : sin (x) - 2 sin^{3} (x)

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

1 \cdot sin (x) = sin (x)

1 sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

2 sin^{2} (x) sin (x) = 2 sin^{3} (x)

2 sin^{2} (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x) + 2 (1 - sin^{2} (x)) sin (x)

Expand

2 sin (x) (1 - sin^{2} (x)) : 2 sin (x) - 2 sin^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

2 \cdot 1 \cdot sin (x) - 2 sin^{2} (x) sin (x) : 2 sin (x) - 2 sin^{3} (x)

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

2 \cdot 1 \cdot sin (x) = 2 sin (x)

2 \cdot 1 sin (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin (x)

2 sin^{2} (x) sin (x) = 2 sin^{3} (x)

2 sin^{2} (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (x)

= 2 sin (x) - 2 sin^{3} (x)

= 2 sin (x) - 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x) + 2 sin (x) - 2 sin^{3} (x)

Simplify

sin (x) - 2 sin^{3} (x) + 2 sin (x) - 2 sin^{3} (x) : - 4 sin^{3} (x) + 3 sin (x)

sin (x) - 2 sin^{3} (x) + 2 sin (x) - 2 sin^{3} (x)

Group like terms

= - 2 sin^{3} (x) - 2 sin^{3} (x) + sin (x) + 2 sin (x)

Add similar elements:

- 2 sin^{3} (x) - 2 sin^{3} (x) = - 4 sin^{3} (x)

= - 4 sin^{3} (x) + sin (x) + 2 sin (x)

Add similar elements:

sin (x) + 2 sin (x) = 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= 3 sin (x) - 4 sin^{3} (x)

= \frac{3 sin ( a ) - 4 sin ^{3} ( a )}{sin ( a )}

Simplify

\frac{3 sin ( a ) - 4 sin ^{3} ( a )}{sin ( a )} : 3 - 4 sin^{2} (a)

\frac{3 sin ( a ) - 4 sin ^{3} ( a )}{sin ( a )}

Factor

3 sin (a) - 4 sin^{3} (a) : sin (a) (3 - 4 sin^{2} (a))

3 sin (a) - 4 sin^{3} (a)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{3} (a) = sin (a) sin^{2} (a)

= 3 sin (a) - 4 sin (a) sin^{2} (a)

Factor out common term

sin (a)

= sin (a) (3 - 4 sin^{2} (a))

= \frac{sin ( a ) ( 3 - 4 sin ^{2} ( a ) )}{sin ( a )}

Cancel the common factor:

sin (a)

= 3 - 4 sin^{2} (a)

= 3 - 4 sin^{2} (a)

= 3 - 4 sin^{2} (a)

Manipulating right side

2 cos (2 a) + 1

Rewrite using trig identities

2 cos (2 a) + 1

Use the Double Angle identity:

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

= 2 (1 - 2 sin^{2} (a)) + 1

Simplify

2 (1 - 2 sin^{2} (a)) + 1 : - 4 sin^{2} (a) + 3

2 (1 - 2 sin^{2} (a)) + 1

Expand

2 (1 - 2 sin^{2} (a)) : 2 - 4 sin^{2} (a)

2 (1 - 2 sin^{2} (a))

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

2 \cdot 1 - 2 \cdot 2 sin^{2} (a) : 2 - 4 sin^{2} (a)

2 \cdot 1 - 2 \cdot 2 sin^{2} (a)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (a)

Multiply the numbers:

2 \cdot 2 = 4

= 2 - 4 sin^{2} (a)

= 2 - 4 sin^{2} (a)

= 2 - 4 sin^{2} (a) + 1

Simplify

2 - 4 sin^{2} (a) + 1 : - 4 sin^{2} (a) + 3

2 - 4 sin^{2} (a) + 1

Group like terms

= - 4 sin^{2} (a) + 2 + 1

Add the numbers:

2 + 1 = 3

= - 4 sin^{2} (a) + 3

= - 4 sin^{2} (a) + 3

= - 4 sin^{2} (a) + 3

= - 4 sin^{2} (a) + 3

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (sin(3a))/(sin(a))=2cos(2a)+1 ?
The answer to whether (sin(3a))/(sin(a))=2cos(2a)+1 is True