Help~STPM Trigonometry

kitwei1219

Prove that for all values of A
(i)cosA + cos3A + cos5A
=cosA(4cos^2 A -3)(4cos^2 A -1)

(ii)sin2A + sin4A + sin6A
=4sinAcosA(1 - 2sin^2 A)(3 - 4sin^2 A)

p/s:what's the simplest way to solve it?

flash

原帖由 kitwei1219 于 29-9-2009 03:49 PM 发表
Prove that for all values of A
(i)cosA + cos3A + cos5A
=cosA(4cos^2 A -3)(4cos^2 A -1)

(ii)sin2A + sin4A + sin6A
=4sinAcosA(1 - 2sin^2 A)(3 - 4sin^2 A)

p/s:what's the simplest way to solve it ...

做第一题，第二题自己试着做。。。。。。。

cos A + cos 3A + cos 5A
= cos 3A + 2 cos 3A cos 2A
= cos 3A (1 + 2 cos 2A)
= (4 cos^3 A - 3 cos A) [ 1 + 2 (2cos^2 A - 1)]
= cos A (4 cos^2 A - 3) (4 cos^2 A - 1)

mathlim

多一个参考，这个不需要cos3A的公式。
cos3A = 4cos³A - 3cosA
这个公式有一个口诀：
块三 等于 四快三 减 三块
注：用福建方言念。

cosA + cos3A + cos5A
= cosA + 2cos4AcosA
= cos A (1 + 2cos4A)
= cosA [1 + 2(2cos²2A - 1)]
= cosA (4cos²2A - 1)
= cosA (2cos2A - 1)(2cos2A + 1)
= cos A (4cos²A - 3)(4 cos²A - 1)

~HeBe~_@

Mathlim也是福建人哦！
不错的口诀。。
我要背起来~^^

kitwei1219

谢谢flash和mathlim

DADDY_MUMMY

请问这是什么意思？

The length of two sides and one non-included acute angle are given. The non-included acute angle is opposite to the shorter sides of two given sides.

外星護法

回复 3# mathlim


    好厉害的口诀~~ 有没有sin3A的 ？