if 5sinA+3cosA=4 find the value of 3sinA-5cosA
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5 Sin A+ 3Cos A= 4 ----> eq 1
3 Sin A- 5 Cos A = x -----> eq 2
squaring and adding both the equations:
(5 sin A+3 cos A)²+ (3sin A-5Cos A)² = 4²+x²
25 sin²A+ 2×3×5 SinA cos A+ 9cos²A + 9²sinA - 2×3×5 sinA cos A + 25 Cos²A = 4² + x²
rearranging :
25 sin ²A + 25Cos²A+9 sin²A +9cos²A+30 sinAcosA-30SinAcosA= 16+x²
(since sin²A+cos²A = 1)
25+9 = 16+x²
x² = 34
x =4.242
3 Sin A- 5 Cos A = x -----> eq 2
squaring and adding both the equations:
(5 sin A+3 cos A)²+ (3sin A-5Cos A)² = 4²+x²
25 sin²A+ 2×3×5 SinA cos A+ 9cos²A + 9²sinA - 2×3×5 sinA cos A + 25 Cos²A = 4² + x²
rearranging :
25 sin ²A + 25Cos²A+9 sin²A +9cos²A+30 sinAcosA-30SinAcosA= 16+x²
(since sin²A+cos²A = 1)
25+9 = 16+x²
x² = 34
x =4.242
aishwarya61:
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