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In the reduction below, I do not understand line 4 and 6. What identities were applied to line 3 and 5 to reach those conclusions? How were those identities introduced?

1 ) $\sin 3a $

2 ) $= \sin(2a +s)$

3 ) $= \sin2a ·\cos a + \cos 2a·\sin a$

4 ) $=(2\sin a·\cos a)\cos a + (\cos^2a - \sin^2a)\sin a$

5 ) $= 2\sin a·\cos²a + \cos²a·\sin a - \sin³a$

6 ) $ = 2\sin a(1-\sin²a) + (1 - \sin²a)\sin a - \sin³a$

7 ) $ = 2\sin a - 2\sin³a + \sin a - \sin³a - \sin³a$

8 ) $= 3\sin a - 4\sin³a$

This is important rewriting all forms of $\sin$ $n·s$ in terms of $\sin$ $s$. All help is greatly appreciated

  • 1
    Pythagorean identity + double angle formulas. – 2017-02-08
  • 0
    Did you find the comment above helpful? – 2017-02-08
  • 1
    Note that: $$\sin(2x)\equiv 2\sin(x)\cos(x)$$ And $$\cos(2x) \equiv \cos^2(x)-\sin^2(x)$$ And $$ \sin^2(x)+\cos^2(x) \equiv 1$$ – 2017-02-08

4 Answers 4

3

Note that for Line $(3)$, we apply the angle addition laws

$$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$

and

$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$

to reveal the double angle formula resepctively for the sine function

$$\begin{align} \sin(2a)&=\sin(a+a)\\\\ &=\sin(a)\cos(a)+\cos(a)+\sin(a)\\\\ &=2\sin(a)\cos(a) \end{align}$$

and cosine function

$$\begin{align} \cos(2a)&=\cos(a+a)\\\\ &=\cos(a)\cos(a)-\sin(a)\sin(a)\\\\ &=\cos^2(a)-\sin^2(a) \end{align}$$

For Line $(4)$ we simply make use of the identity $\sin^2(x)+\cos^2(x)=1$.

1

In line 3, the identity $\sin(2a)=2\sin(a)\cos(a)$ was used.

In Line 6, the identity $\cos^2(a)=1-\sin^2(a)$ was used

1

Following are the formula's used above -

1.) $\sin(a + b) = \sin a \cos b + \cos a \sin b$

2.) $\sin 2a = 2 \sin a \cos a$

3.) $\cos 2a = \cos^2 a - \sin^2 a$

4.) $\cos^2 a = 1 - \sin^2 a$

0

The angle addition formula is applied to $\sin(2a) = \sin(a+a)$ and $\cos(2a) = \cos(a+a)$.

In line 6 the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$ is used.