Trigonometry (hard)

[匿名]

2011-10-17 3:19 am

1.

(sinx)^2 + sin2x = m + 2(cosx)^2

Express the equation in terms of sin2x and cos 2x

2.

A, B,C are interior angles of triangle ABC. Prove

1 - cos(A + B)cos(A - B) + (sinC)^2 = 2 + 2cosAcosBcosC

回答 (2)

用戶2053

2011-10-17 4:09 am

✔ 最佳答案

1)sin²x + sin2x = m + 2cos²x(1 - cos2x)/2 + sin2x = m + (cos2x + 1)1 - cos2x + 2sin2x = 2m + 2cos2x - 22m = 2sin2x - 3cos2x + 3m = sin2x - (3/2)cos2x + 3/2
2)- cos(A+B) cos(A -B) + sin²C= - cos(A+B) cos(A -B) + 1 - cos²C= - cos(A+B) cos(A -B) + 1 - cos²(A+B)= 1 - cos(A+B) ( cos(A -B) + cos(A+B) )= 1 - cos(A+B) ( cosAcosB + sinAsinB + cosAcosB - sinAsinB ) = 1 - cos(A+B) (2cosAcosB)= 1 + 2 cosA cosB cosC

2011-10-16 20:11:40 補充：
Corrections :

1)

sin²x + sin2x = m + 2cos²x

(1 - cos2x)/2 + sin2x = m + (cos2x + 1)

1 - cos2x + 2sin2x = 2m + 2cos2x + 2

2m = 2sin2x - 3cos2x - 1

m = sin2x - (3/2)cos2x - 1/2

2011-10-16 20:13:15 補充：
- cos(A + B)cos(A - B) + (sinC)^2

= 1 + 2cosAcosBcosC instead of 2 + 2cosAcosBcosC.

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CHI FUNG

2011-10-17 4:58 am

did you see the ''1'' at the front of this question, xd

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