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Question: If \(\text{sinA}\ \text{+}\ \text{si}{{\text{n}}^{2}}\text{A+si}{{\text{n}}^{3}}\text{A=1,}\) then f...

If sinA + sin2A+sin3A=1,\text{sinA}\ \text{+}\ \text{si}{{\text{n}}^{2}}\text{A+si}{{\text{n}}^{3}}\text{A=1,} then find the value of cos⁡6A − 4cos⁡4A + 8cos2A{{\cos }^{6}}\text{A}\ -\ 4{{\cos }^{4}}\text{A}\ \text{+}\ \text{8co}{{\text{s}}^{2}}\text{A}is:-
A). 1.
B). 2.
C). 3.
D). 4

Explanation

Solution

Try using the equation given to us,sinA + sin2A+sin3A=1,\text{sinA}\ \text{+}\ \text{si}{{\text{n}}^{2}}\text{A+si}{{\text{n}}^{3}}\text{A=1,}you can use the basic trigonometry property, sin⁡2A + cos2A = 1,{{\sin }^{2}}\text{A}\ \text{+}\ \text{co}{{\text{s}}^{2}}\text{A}\ \text{=}\ \text{1,} which will be helpful in getting to the solution.

Complete step-by-step answer:
We need to find the value of cos⁡6A - 4cos4A + 8cos2A.{{\cos }^{6}}\text{A}\ \text{- 4co}{{\text{s}}^{4}}\text{A}\ \text{+}\ \text{8co}{{\text{s}}^{2}}\text{A}\text{.} Let us use the equation given to us.
⇒ sin⁡A + sin2A + sin3A =1\Rightarrow \ \sin \text{A}\ \text{+}\ \text{si}{{\text{n}}^{2}}\text{A}\ \text{+}\ \text{si}{{\text{n}}^{3}}\text{A}\ \text{=1}
Taking sin⁡2A{{\sin }^{2}}\text{A} to the right hand side,
⇒ sinA + sin3A = 1 - sin2A\Rightarrow \ \text{sinA}\ \text{+}\ \text{si}{{\text{n}}^{\text{3}}}\text{A}\ \text{=}\ \text{1}\ \text{-}\ \text{si}{{\text{n}}^{\text{2}}}\text{A}
⇒ sinA(1+sin2A)= cos2A\Rightarrow \ \text{sinA}\left( 1+\text{si}{{\text{n}}^{\text{2}}}\text{A} \right)=\ \text{co}{{\text{s}}^{\text{2}}}\text{A}
Here we can substitute the value of sin2A   as    1- cos2A\text{si}{{\text{n}}^{\text{2}}}\text{A}\ \ \ \text{as}\ \ \ \ \text{1-}\ \text{co}{{\text{s}}^{\text{2}}}\text{A}
Therefore we will get,
sinA (1+1−cos2A)= cos2A\text{sinA}\ \left( 1+1-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)=\ \text{co}{{\text{s}}^{\text{2}}}\text{A}
sinA(2−cos2A)=cos⁡2(A)\text{sinA}\left( 2-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)={{\cos }^{2}}\left( \text{A} \right)
Now squatting on both the sides,
sin2A(2−cos2A)2 = cos⁡4A\text{si}{{\text{n}}^{\text{2}}}\text{A}{{\left( 2-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)}^{2}}\ =\ {{\cos }^{4}}A
Again we can replace sin2A by (1−cos2A)\text{si}{{\text{n}}^{\text{2}}}\text{A}\ \text{by}\ \left( 1-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)
⇒ (1−cos2A)(4+cos4A - 4cos2A)= cos4A\Rightarrow \ \left( 1-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)\left( 4+\text{co}{{\text{s}}^{\text{4}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A} \right)=\ \text{co}{{\text{s}}^{\text{4}}}\text{A}
⇒ 4+cos4A - 4cos2A - 4cos2A - cos6A + 4cosA4A = cos4A\Rightarrow \ \text{4+co}{{\text{s}}^{\text{4}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{-}\ \text{co}{{\text{s}}^{\text{6}}}\text{A}\ \text{+}\ \text{4cos}{{\text{A}}^{\text{4}}}\text{A}\ \text{=}\ \text{co}{{\text{s}}^{\text{4}}}\text{A}
Taking all the cos⁡\cos terms on are side,
cos4A - cos4A +4cos2A + 4cos2A + cos6A -4cos4A = 4\text{co}{{\text{s}}^{\text{4}}}\text{A}\ \text{-}\ \text{co}{{\text{s}}^{\text{4}}}\text{A}\ \text{+4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{+}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{+}\ \text{co}{{\text{s}}^{\text{6}}}\text{A}\ \text{-4co}{{\text{s}}^{\text{4}}}\text{A}\ \text{=}\ \text{4}
⇒ cos6A - 4cos4A + 8cos2A = 4\Rightarrow \ \text{co}{{\text{s}}^{\text{6}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{4}}}\text{A}\ \text{+}\ \text{8co}{{\text{s}}^{\text{2}}}\text{A}\ \text{=}\ \text{4}

Hence, our answer will be 4.

Note: In such questions rearrangement of the given equation always leads us to the desired equation. Carefully verify each step before moving into the next step. Additionally some basic trigonometric ideates to keep in mind would be,
sin⁡2θ +cos⁡2θ=1{{\sin }^{2}}\theta \ +{{\cos }^{2}}\theta =1
tan⁡2θ+1=sec⁡2θ{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta
cot⁡2θ+1=cosec2θ{{\cot }^{2}}\theta +1=\text{cose}{{\text{c}}^{\text{2}}}\theta .