
What is $\sin \left( x \right)+\cos \left( x \right)$ in terms of sine?
Answer
524.4k+ views
Hint: We solve this question by using a trigonometric identity given by ${{\sin }^{2}}x+{{\cos}^{2}}x=1.$ We rearrange these terms and simplify to obtain the value of $\cos x$ and substitute this in the equation given in the question. By doing so, we get the expression in terms of sine.
Complete step-by-step solution:
In order to answer this question, let us state a basic trigonometric identity for sin and cos. This is represented as follow:
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1$
We need to find out the expression for $\cos x$ so that we can substitute this for the cos term in the question and obtain the answer in terms of sine only.
For this, we subtract both sides of the equation by ${{\sin }^{2}}x,$
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x=1-{{\sin }^{2}}x$
$\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x$
Taking square root on both sides of the equation,
$\Rightarrow \sqrt{{{\cos }^{2}}x}=\sqrt{1-{{\sin }^{2}}x}$
The root and square cancel out on the left-hand side of the equation to yield,
$\Rightarrow \cos x=\sqrt{1-{{\sin }^{2}}x}\ldots \left( 1 \right)$
Now, we consider the expression given in the question.
$\Rightarrow \sin \left( x \right)+\cos \left( x \right)$
We replace the value of $\cos x$ from equation 1 in the above equation.
$\Rightarrow \sin \left( x \right)+\sqrt{1-{{\sin }^{2}}x}$
Hence, we have represented the value of the expression $\sin \left( x \right)+\cos \left( x \right)$ in terms if sine.
We can also do this by using the concept of complementary identity given by,
$\Rightarrow \cos x=\sin \left( \dfrac{\pi }{2}-x \right)$
Substituting this in equation 1,
$\Rightarrow \sin \left( x \right)+\sin \left( \dfrac{\pi }{2}-x \right)$
Hence, we have two ways to represent the expression $\sin \left( x \right)+\cos \left( x \right)$ in terms if sine given by $\sin \left( x \right)+\sqrt{1-{{\sin }^{2}}x}$ and $\sin \left( x \right)+\sin \left( \dfrac{\pi }{2}-x \right).$ .
Note: We need to know the basic concepts of trigonometric functions in order to solve this question. It is to be noted that there is not just one way of representing this expression in terms of sine. It can be done in more ways and depends on the identity we decide to use.
Complete step-by-step solution:
In order to answer this question, let us state a basic trigonometric identity for sin and cos. This is represented as follow:
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1$
We need to find out the expression for $\cos x$ so that we can substitute this for the cos term in the question and obtain the answer in terms of sine only.
For this, we subtract both sides of the equation by ${{\sin }^{2}}x,$
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x=1-{{\sin }^{2}}x$
$\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x$
Taking square root on both sides of the equation,
$\Rightarrow \sqrt{{{\cos }^{2}}x}=\sqrt{1-{{\sin }^{2}}x}$
The root and square cancel out on the left-hand side of the equation to yield,
$\Rightarrow \cos x=\sqrt{1-{{\sin }^{2}}x}\ldots \left( 1 \right)$
Now, we consider the expression given in the question.
$\Rightarrow \sin \left( x \right)+\cos \left( x \right)$
We replace the value of $\cos x$ from equation 1 in the above equation.
$\Rightarrow \sin \left( x \right)+\sqrt{1-{{\sin }^{2}}x}$
Hence, we have represented the value of the expression $\sin \left( x \right)+\cos \left( x \right)$ in terms if sine.
We can also do this by using the concept of complementary identity given by,
$\Rightarrow \cos x=\sin \left( \dfrac{\pi }{2}-x \right)$
Substituting this in equation 1,
$\Rightarrow \sin \left( x \right)+\sin \left( \dfrac{\pi }{2}-x \right)$
Hence, we have two ways to represent the expression $\sin \left( x \right)+\cos \left( x \right)$ in terms if sine given by $\sin \left( x \right)+\sqrt{1-{{\sin }^{2}}x}$ and $\sin \left( x \right)+\sin \left( \dfrac{\pi }{2}-x \right).$ .
Note: We need to know the basic concepts of trigonometric functions in order to solve this question. It is to be noted that there is not just one way of representing this expression in terms of sine. It can be done in more ways and depends on the identity we decide to use.
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