Question

IF sinA + sin²A+ sin³A =1 prove that cos⁶A +4cos⁴A +8cos²A =4

Answer 1

If

$$\sin A + \sin^ {2} A + \sin^ {3} A = 1$$

prove that

$$\cos^ {6} A - 4 \cos^ {4} A + 8 \cos^ {2} A = 4.$$

**Solution:**

We’ll use the following trigonometric identity

$$\sin^ {2} A = 1 - \cos^ {2} A.$$

Thus we have

$$\sin A + \sin^ {2} A + \sin^ {3} A = 1,$$ $$\sin A + 1 - \cos^ {2} A + \sin^ {3} A = 1,$$ $$\sin A + \sin^ {3} A - \cos^ {2} A = 0,$$ $$\sin A (1 + \sin^ {2} A) = \cos^ {2} A,$$ $$\sin A (2 - \cos^ {2} A) = \cos^ {2} A,$$ $$\left(\sin A (2 - \cos^ {2} A)\right) ^ {2} = (\cos^ {2} A) ^ {2},$$ $$\sin^ {2} A (2 - \cos^ {2} A) ^ {2} = \cos^ {4} A,$$ $$(1 - \cos^ {2} A) (4 - 4 \cos^ {2} A + \cos^ {4} A) = \cos^ {4} A,$$ $$4 - 4 \cos^ {2} A + \cos^ {4} A - 4 \cos^ {2} A + 4 \cos^ {4} A - \cos^ {6} A = \cos^ {4} A,$$ $$4 = 8 \cos^ {2} A - 4 \cos^ {4} A + \cos^ {6} A,$$ $$\cos^ {6} A - 4 \cos^ {4} A + 8 \cos^ {2} A = 4$$