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標題:

A maths Trigonometry

發問:

1. In triangle ABC, if sin C= (sin A+sin B)/(cos A+cos B), prove that triangle ABC is a right triangle. 2. Find the maximum value and minimum value of 3/(4(cos x)^2+5)-1 更新: The answer of question 2 is -2/5 and 2/5, How can I get the answer??

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最佳解答:

1 In triangle ABC, if sin C= (sin A+sin B)/(cos A+cos B), prove that triangle ABC is a right triangle. SOLUTION sin A + sin B sin C = -------------------- --- cos A + cos B sin C=(sin A + sin B)/(cos A + cos B) sin C=(sin (A+B)/2 cos(A- B)/2)/(cos (A+B)/2 cos (A-B)/2) sin C=(sin (90-C/2) cos(A- B)/2)/(cos (90-C/2) cos (A-B)/2) sin C=cos (C/2) /sin (C/2) 2(sinC/2)(cosC/2)=co s (C/2) /sin (C/2) sin^2(C/2)=1/2 So C/2=45 C=90 ?ABC is a right-angled triangle and C is the right-angled Note: (A+B)/2=(180-C)/2=90-C/2 2 Find the maximum value and minimum value of 3/(4(cos x)^2+5)-1 SOLUTION Because -1<=cos x<=1 0<=(cos x)^2<=1 0<=4(cos x)^2<=4 5<=4(cos x)^2+5<=9 1/9<=1/(4(cos x)^2+5)<=1/5 3/9<=3/(4(cos x)^2+5)<=3/5 1/3<=3/(4(cos x)^2+5)<=3/5 -2/3<=3/(4(cos x)^2+5)-1<=-2/5 So the maximum value is -2/5, the minimum value is -2/3 2007-04-08 15:27:12 補充： http://hk.knowledge.yahoo.com/question/?qid=7007022600167佢同我既答案都一樣bor所以應該係書既答案錯

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