高一三角恒等

题目

高一三角恒等
1) 证明 tan^2 x + 1/(tan^2 x) = [2(3+2cos4x)]/(1-cos4x)
2) (1+tan22)(1+tan23)(1+tan24)(1+tan25)=?

答案

(1)
证明:
tan^2(x)+1/[tan^2(x)]
=[sin^2(x)/cos^2(x)]+[cos^2(x)/sin^2(x)]
={[sin^4(x)+cos^4(x)]/[sin^2(x)cos^2(x)]
由于:
分子：sin^4(x)+cos^4(x)
=[sin^2(x)]^2+[cos^2(x)]^2
=[(1-cos2x)/2]^2+[(1+cos2x)/2]^2
=[1/4+1/4cos^2(2x)-1/2cos2x]
+[1/4+1/4cos^2(2x)+1/2cos2x]
=1/2+1/2cos^2(2x)
=1/2+1/2*[(1+cos4x)/2]
=3/4+1/4*cos4x
=(3+cos4x)/4
分母:sin^2(x)cos^2(x)
=[(1-cos2x)/2]*[(1+cos2x)/2]
=[1-cos^2(2x)]/4
={1-[(1+cos4x)/2]}/4
=(1-cos4x)/8
则:分子/分母
=[(3+cos4x)/4]/[(1-cos4x)/8]
=[2(3+cos4x)]/(1-cos4x)
即命题得证
(2)
(1+tan22)(1+tan23)(1+tan24)(1+tan25)
=[(1+tan22)(1+tan23)][(1+tan24)(1+tan25)]
=[1+tan22*tan23+(tan22+tan23)]*[1+tan24*tan25+(tan24+tan25)]
由于
tan45
=tan(22+23)
=(tan22+tan23)/(1-tan22*tan23)
=1
所以
tan22+tan23
=tan45*(1-tan22*tan23)
=1-tan22*tan23
则:
(1+tan22)(1+tan23)(1+tan24)(1+tan25)
=[1+tan22*tan23+(tan22+tan23)]*[1+tan24*tan25+(tan24+tan25)]
=[1+tan22*tan23+(1-tan22*tan23)]*[1+tan24*tan25+(tan24+tan25)]
=2+2tan24*tan25+2(tan24+tan25)