Chứng minh đẳng thức lượng giác lớp 10 

BÀI 1: $\sin ^4a+\cos ^4a=\dfrac34+\dfrac14\cos 4a\left(1\right)$

Hướng dẫn: 

Bài giải:

$a^2+b^2=\left({a+b}\right)^2-2ab$

$VT=\left({{\sin }^2a}\right)^2+\left({{\cos }^2a}\right)^2$

$=\left({{\sin }^2a+{\cos }^2a}\right)^2-2\sin ^2a.\cos ^2a$

$=1-2\left({\sin a.\cos a}\right)^2$

$=1-2.\left({\dfrac12\sin 2a}\right)^2$ $=1-2.\dfrac14\sin ^22a$

$=1-\dfrac12.\dfrac{1-\cos 4a}2$ $=\dfrac34+\dfrac14\cos 4a=VP$ (đpcm)

BÀI 2: $sin ^6a+\cos ^6a=\dfrac58+\dfrac38\cos 4a\left(2\right)$

Hướng dẫn: 

Bài giải:

Ta có: $a^3+b^3=\left({a+b}\right)^3-3ab\left({a+b}\right)$

$VT=\left({{\sin }^2a}\right)^3+\left({{\cos }^2a}\right)^3$

=$\left({{\sin }^2a+{\cos }^2a}\right)^3-3\sin ^2a.\cos ^2a\left({{\sin }^2a+{\cos }^2a}\right)$

=$1-3\left({\sin{a}.cosa}\right)^2$ $=1-3\left({\dfrac12\sin 2a}\right)^2$

$=1-3.\dfrac14\sin ^22a$ $=1-\dfrac34\left({\dfrac{1-\cos 4a}2}\right)$

$=\dfrac58+\dfrac38\cos 4a=VP$ (đpcm)

BÀI 3: $1-\sin a=2\sin ^2\left({\dfrac\pi 4-\dfrac{a}{2}}\right)\left(3\right)$

Hướng dẫn:

Bài giải:

Ta có: $\sin ^2a=\dfrac{1-\cos 2a}2$

Do đó: $VP=1-\cos \left({\dfrac{\pi}{2}-a}\right)$ $=1-\sin a=VT$ (đpcm)

BÀI 4: $\sin a\cos ^3a-\cos a\sin ^3a=\dfrac14\sin 4a\left(4\right)$

Bài giải:

Ta có: $VT=\sin a.\cos a\left({{\cos }^2a-{\sin }^2a}\right)$

$=\dfrac12\sin 2a.\cos 2a$ =$\dfrac12.\dfrac12\sin 4a=\dfrac14\sin 4a$

=VP