Prove that
1 - cos2 A = sin2 A
Answers
Answer:
cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1 xggfsin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1
\tan(2x) = \dfrac{2 \tan(x)}{1 - \tan^2(x)}tan(2x)=
1−tan
2
(x)
2tan(x)
Half-Angle Identities
\sin\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(x)}{2}}sin(
2
x
)=±
2
1−cos(x)
\cos\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 + \cos(x)}{2}}cos(
2
x
)=±
2
1+cos(x)
\tan\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(x)}{1 + \cos(x)}}tan(
2
x
)=±
1+cos(x)
1−cos(x)
= \dfrac{1 - \cos(x)}{\sin(x)}=
sin(x)
1−cos(x)
= \dfrac{\sin(x)}{1 + \cos(x)}=
1+cos(x)
sin(x)
The above identities can be re-stated by squaring each side and doubling all of the angle measures. The results are as follows:
\sin^2(x) = \frac{1}{2} \big[1 - \cos(2x)\big]sin
2
(x)=
2
1
[1−cos(2x)]
\cos^2(x) = \frac{1}{2} \big[1 + \cos(2x)\big]cos
2
(x)=
2
1
[1+cos(2x)]
\tan^2(x) = \dfrac{1 - \cos(2x)}{1 + \cos(2x)}tan
2
(x)=
1+cos(2x)
1−cos(2x)