Question

If we have an trigonometric expression as \[\sin x+{{\sin }^{2}}x=1\], then the value of \[{{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x+1\] is equal to
(A). 0
(B). 1
(D). 3
(D). 2

Answer 1

Hint: Use \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] when an equation is given in terms of sine, cosine, tangent. We must use any of the trigonometric identities to make the equation solvable. There are many inter-relations between sine, cosine, tan, secant these are inter-relations are called as identities. Whenever you see conditions such that \[\theta \in R\], that means inequality is true for all angles. So, directly think of identity which will make you work easy.

Complete step-by-step solution -
An equality with sine, cosine, or tangent in them is called a trigonometric equation. These are solved by some inter-relations known before-hand.
All the inter-relations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for proof. These are the main and crucial steps to take us nearer to the result.
Given trigonometric relation, in terms of sin is:
 \[\sin x+{{\sin }^{2}}x=1\]
By subtracting with term \[{{\sin }^{2}}x\] on both sides of above equation, we get:
\[\Rightarrow \]\[\sin x=1-{{\sin }^{2}}x\]
By general trigonometry, we can say the identity of sin, cos as:
\[\Rightarrow \]\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
By substituting this in above equation, we get it as:
\[\Rightarrow \]\[\sin x={{\cos }^{2}}x\Rightarrow {{\cos }^{2}}x=\sin x\] --- (1)
By squaring on both sides of a above equation, we get it as:
\[\Rightarrow {{\cos }^{4}}x={{\sin }^{2}}x\] ------- (2)
Given expression for which we need to find the value is:
\[\Rightarrow {{\cos }^{12}}x+3{{\cos }^{10}}x+3{{\cos }^{8}}x+{{\cos }^{6}}x+1\]
We can write this expression, with using (1), (2) in simpler way:
\[\Rightarrow {{\left( {{\cos }^{4}}x \right)}^{3}}+3{{\left( {{\cos }^{4}}x \right)}^{2}}{{\cos }^{2}}x+3\left( {{\cos }^{4}}x \right){{\left( {{\cos }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{3}}+1\]
By general algebraic identity we can say that a, b relation:
\[\Rightarrow {{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\]
By substituting this into our expression it turns into:
\[\Rightarrow {{\left( {{\cos }^{4}}x+{{\cos }^{2}}x \right)}^{3}}+1\]
By substituting equation (2) in this equation, we get it as:
\[\Rightarrow {{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}+1\]
By using general identity of sin, \[\cos x\] which is given by:
\[\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1\]
By substituting this our expression turns into: \[{{1}^{3}}+1\].
By simplifying we can say the value to be 2.
So, the value of the given expression (in terms of \[\cos x\]) is 2.
Therefore option (d) is the correct answer for this condition.

Note: Generally students take \[\sqrt{{{\sin }^{2}}x}\] as \[\sin x\]. By the limits are \[\dfrac{-\pi }{2}\] to \[\dfrac{\pi }{2}\]. So, you must observe sin will be both positive and also negative in this region. So, you must divide region as \[\sqrt{{{\sin }^{2}}x}\] value changes according to region. Alternately you can find P, Q values till end separately and above the integration only once if you find them separately then you must do the same integration twice to get to the same result.