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Question: Write the value of \[\sin (45^\circ + \theta ) - \cos (45^\circ - \theta )\]...

Write the value of sin⁡(45∘+θ)−cos⁡(45∘−θ)\sin (45^\circ + \theta ) - \cos (45^\circ - \theta )

Explanation

Solution

Hint : First of all, we will observe the given expression and will apply the identities accordingly. After applying the identities, simplify and place the values for the angle 45∘45^\circ and then simplify for the resultant value.

Complete step-by-step answer :
Take the given expression –
sin⁡(45∘+θ)−cos⁡(45∘−θ)\sin (45^\circ + \theta ) - \cos (45^\circ - \theta )
We can observe that we can apply the standard general formula for sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta and other identity as cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta in the above expression.
=[sin⁡45∘cos⁡θ+cos⁡45∘sin⁡θ]−[cos⁡45∘cos⁡θ+sin⁡45∘sin⁡θ]= [\sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta ] - [\cos 45^\circ \cos \theta + \sin 45^\circ \sin \theta ]
Remember when there is a negative sign outside the bracket, the sign of all the terms inside the bracket also changes. A positive term becomes negative and vice versa.
=[sin⁡45∘cos⁡θ+cos⁡45∘sin⁡θ−cos⁡45∘cos⁡θ−sin⁡45∘sin⁡θ]= [\sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta - \cos 45^\circ \cos \theta - \sin 45^\circ \sin \theta ]
Place the values of sine and cosine angles in the above equation, as we know that sin⁡45∘=cos⁡45∘=12\sin 45^\circ = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }}
=[12cos⁡θ+12sin⁡θ−12cos⁡θ−12sin⁡θ]= [\dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta - \dfrac{1}{{\sqrt 2 }}\cos \theta - \dfrac{1}{{\sqrt 2 }}\sin \theta ]
Take out common multiples from all the terms in the above expression-
=12[cos⁡θ+sin⁡θ−cos⁡θ−sin⁡θ]= \dfrac{1}{{\sqrt 2 }}[\cos \theta + \sin \theta - \cos \theta - \sin \theta ]
Make the pairs of the like terms in the above expression –
=12[cos⁡θ−cos⁡θ‾+sin⁡θ−sin⁡θ‾]= \dfrac{1}{{\sqrt 2 }}[\underline {\cos \theta - \cos \theta } + \underline {\sin \theta - \sin \theta } ]
The terms with equal values and opposite sign cancel each other.
=12[0]= \dfrac{1}{{\sqrt 2 }}[0]
Anything multiplied with zero, gives the resultant value as the zero.
=0= 0
Hence, the value of sin⁡(45∘+θ)−cos⁡(45∘−θ)=0\sin (45^\circ + \theta ) - \cos (45^\circ - \theta ) = 0
So, the correct answer is “0”.

Note : Remember the All STC rule, it is also known as ASTC rule in the geometry. It states that all the trigonometric ratios in the first quadrant ( 0∘  to 90∘0^\circ \;{\text{to 90}}^\circ ) are positive, sine and cosec are positive in the second quadrant ( 90∘ to 180∘90^\circ {\text{ to 180}}^\circ ), tan and cot are positive in the third quadrant ( 180∘  to 270∘180^\circ \;{\text{to 270}}^\circ ) and sin and cosec are positive in the fourth quadrant ( 270∘ to 360∘270^\circ {\text{ to 360}}^\circ ).