Question

Sino + cos 0 = 17/13
then Sino-coso =?
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Answer 1

Answer:

Instead of theta(0) I'll be typing A, because it is easier☺

sinA + cosA = 17/13

(sinA + cosA)² = 289/169

sin²A + cos²A + 2sinAcosA = 289/169

2sinAcosA = 289/169 - 1 =

2sinAcosA = 120/169

sinA - cosA = √[(sinA - cosA)²]

= √[sin²A + cos²A -2sinAcosA]

= √[1 - 120/169]

= √[49/169]

= √49/√169

= 7/13

sinA - cosA = 7/13