if sinA+sin2A=1 then evaluate the expression cos2A+cos4A.
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Given, sin A + 2 cos A = 1
Squaring both sides, we get,
sin2A + 4 cos2A + 4sinA cosA = 1
4 cos2A + 4sinA cosA = 1 - sin2A = cos2A
3 cos2A + 4sinA cosA = 0 ... (i)
Now, (2sinA - cosA)2 = 4sin2A + cos2A - 4sinA cosA
= 4sin2A + cos2A + 3cos2A [Using (i)]
= 4 (sin2A + cos2A) = 4
Thus, 2sinA - cosA = 2
Hence, proved.
Squaring both sides, we get,
sin2A + 4 cos2A + 4sinA cosA = 1
4 cos2A + 4sinA cosA = 1 - sin2A = cos2A
3 cos2A + 4sinA cosA = 0 ... (i)
Now, (2sinA - cosA)2 = 4sin2A + cos2A - 4sinA cosA
= 4sin2A + cos2A + 3cos2A [Using (i)]
= 4 (sin2A + cos2A) = 4
Thus, 2sinA - cosA = 2
Hence, proved.
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