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10 When 0° <A <90, solve the following equations:(i) sín 3A = cos 2A(ii) tan 5A-cot A. |
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Answer» sin3A = cos2Asin3A = sin(90- 2A)Cancelling sin from both sides.We get,3A = 90 - 2A5A = 90therefore A = 90/5 = 18 degrees. If you ask why did i change cosA on the RHS to sin(90-2A)....I used the identity cosA = sin(90-A).If you ask why did i cancel sin then i'll say sin3A is not sin of angle 3A and it is not to be confused with sinA because sinA is sin of angle A and sin3A is three times sin of angle A or Sin * 3A.. |
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