Question

Evaluate the value of \[\sin {{45}^{\circ }}+\cos {{45}^{\circ }}\].

Answer 1

Hint:We are given with \[\sin {{45}^{\circ }}+\cos {{45}^{\circ }}\]. Since these are the basic trigonometric terms, we know the direct values of the two terms. We can obtain the value of \[\sin {{45}^{\circ }}+\cos {{45}^{\circ }}\] by simply substituting the value the terms and adding them gives us the required value.

Complete step-by-step solution:
Now let us have a brief regarding the trigonometric ratios. The six basic trigonometric ratios are sine, cosine, tangent, secant, cosecant and cotangent. These are called ratios because they are expressed in the ratios of sides of a right angled triangle. Trigonometric identities are those which are true for all values of the variables. We can convert the degrees to radians and radian to degrees. We can convert any radian to degree just by multiplying the given radian with \[\dfrac{{{180}^{\circ }}}{\pi }\].
Now let us find the value of \[\sin {{45}^{\circ }}+\cos {{45}^{\circ }}\]
We know that,
\[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\] and
\[\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\]
Now upon substituting the values of the respective functions, we get
\[\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}\]
We can express \[\dfrac{2}{\sqrt{2}}\] also as,
\[\dfrac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}}=\sqrt{2}\]
\[\therefore \] The value of \[\sin {{45}^{\circ }}+\cos {{45}^{\circ }}\] is \[\sqrt{2}\].

Note: We must have a note that if the angles are given from any of the given quadrants then we can reduce the angle to the equivalent of the first quadrant by changing the signs and the trigonometric ratios. We can apply trigonometry functions in everyday life in astronomy, seismology, navigation , electronics etc. we can also find the height of the buildings, mountains trees etc with the help of \[\tan \] function.
The graph of \[\sin {{45}^{\circ }}+\cos {{45}^{\circ }}\] is shown below.