Question

write cos2x as sinx
please help with this ​

Answer 1

Answer:

$\cos(2\, x) = 1 - 2\, (\sin(x))^2$.

Step-by-step explanation:

Angle sum identity for cosine: $\cos(a + b) = \cos(a) \, \cos(b) - \sin(a) \, \sin(b)$.

Pythagorean identity: $(\cos(a))^{2} + (\sin(a))^{2} = 1$ for all real $a$.

Subtract $(\cos(x))^{2}$ from both sides of the Pythagorean identity to obtain: $(\sin(a))^{2} = 1 - (\cos(a))^{2}$.

Apply angle sum identity to rewrite $\cos(2\, x)$.

$\begin{aligned}&\cos(2\, x)\\ &= \cos(x + x) \\ &= \cos(x) \, \cos(x) - \sin(x)\, \sin(x) \\ &= (\cos(x))^{2} + (\sin(x))^{2}\end{aligned}$.

$(\sin(a))^{2} = 1 - (\cos(a))^{2}$ follows from the Pythagorean identity. Hence, it would be possible to replace the $(\cos(x))^{2}$ in the previous expression with $(1 - (\sin(x))^{2})$.

$\begin{aligned}&(\cos(x))^{2} - (\sin(x))^{2}\\ &= \left[1 - (\sin(x))^{2}\right] - (\sin(x))^{2} \\ &= 1 - 2\, (\sin(x))^{2} \end{aligned}$.

Conclusion:

$\begin{aligned}&\cos(2\, x) \\ &= (\cos(x))^{2} + (\sin(x))^{2} \\ &=1 - 2\, (\sin(x))^{2}\end{aligned}$