Question
Question: Simplify: \(\dfrac{{{\sin }^{3}}\theta +{{\cos }^{3}}\theta }{\sin \theta +\cos \theta }+\sin \theta...
Simplify: sinθ+cosθsin3θ+cos3θ+sinθcosθ?
Solution
To simplify the given trigonometric expression, we are going to use the property of the algebraic expression which is as follows: a3+b3=(a+b)(a2+b2−ab). This algebraic identity we are going to use in sin3θ+cos3θ. After applying this algebraic identity, (sinθ+cosθ) will get cancelled out from the numerator and the denominator in the given expression. After that using simple algebra, we can solve the given expression.
Complete step by step answer:
The trigonometric expression given in the above problem is as follows:
sinθ+cosθsin3θ+cos3θ+sinθcosθ
To solve the above expression, we are going to use the following algebraic identity which is equal to:
a3+b3=(a+b)(a2+b2−ab)
We are going to apply the above identity in sin3θ+cos3θ. So, substituting a=sinθ&b=cosθ in the above equation we get,
sin3θ+cos3θ=(sinθ+cosθ)(sin2θ+cos2θ−sinθcosθ)
Now, applying the above equation in the given expression and we get,
sinθ+cosθ(sinθ+cosθ)(sin2θ+cos2θ−sinθcosθ)+sinθcosθ
As you can see that in the above expression, (sinθ+cosθ) is common in the numerator and the denominator so (sinθ+cosθ) will get cancelled out and we are left with:
(sin2θ+cos2θ−sinθcosθ)+sinθcosθ
Now, in the above expression, you can see that sinθcosθ will get cancelled out and the above expression will look like:
sin2θ+cos2θ
We also know the trigonometric identity which says that addition of square of sine and addition of square of cosine will give an answer as 1. The mathematical form of the above trigonometric identity will look like:
sin2θ+cos2θ=1
Hence, the simplification of the expression given in the above problem is 1.
Note: To solve the above problem, you must know the algebraic identity of a3+b3=(a+b)(a2+b2−ab) and also the trigonometric identity which says that square of sine and square of cosine is 1. Failure to remember any of the identities inhibits you to solve the question further.