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[F. 3 Maths] Trigonometric Relations (Only 3 questions!)

發問:

[F. 3 Maths] Trigonometric Relations (Only 3 questions!) Solve for x in the following equations. 1. sin ( 4x + 60' ) = cos 6x 2. sin ( 3x + 11' ) = cos ( 2x - 6' ) 3. 1/(tan 4x) = (sin 5x)/(cos 5x) Remarks: 1. ' = degree(s) 2. / = over

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• 幫我找一些形容藝術的詞語(爆急)
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• 大埔邊度可以睇歐聯決賽-

• 幫我找一些形容藝術的詞語(爆急)
• 我想??死
• 大埔邊度可以睇歐聯決賽-

 

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1. sin ( 4x + 60' ) = cos 6x =cos [90-(4x+60)] = cos 6x Therefore [90-(4x+60)] = 6x 90-4x-60 = 6x 30-4x = 6x 30 = 10x x = 3 2. sin ( 3x + 11' ) = cos ( 2x - 6' ) =cos[90- (3x+11)] = cos(2x-6) Therefore 90- (3x+11)= 2x-6 90-3x-11 = 2x-6 79-3x = 2x-6 79+6 = 2x+3x 85 = 5x x =17 3. 1/(tan 4x) = (sin 5x)/(cos 5x) = tan (90-4x)= tan 5x Therefore 90-4x= 5x 90 = 9x x =10

其他解答:

1. sin ( 4x + 60' ) = cos 6x cos (90 - 4x+60')=cos 6x 90- 4x-60=6x 10x=30 x=3 2. sin ( 3x + 11' ) = cos ( 2x - 6' ) cos (90-3x+11')=cos ( 2x - 6' ) 90-3x-11'=2x-6' 5x=85 x=17 3. 1/(tan 4x) = (sin 5x)/(cos 5x) 1/tan4x=tan5x tan(90-4x)=tan 5x 90-4x=5x 9x=90 x=10'