
Write the value of \[\sin (45^\circ + \theta ) - \cos (45^\circ - \theta )\]
Answer
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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^\circ $ and then simplify for the resultant value.
Complete step-by-step answer:
Take the given expression –
\[\sin (45^\circ + \theta ) - \cos (45^\circ - \theta )\]
We can observe that we can apply the standard general formula for $ \sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $ and other identity as $ \cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $ in the above expression.
$ = [\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^\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^\circ = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }} $
$ = [\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-
$ = \dfrac{1}{{\sqrt 2 }}[\cos \theta + \sin \theta - \cos \theta - \sin \theta ] $
Make the pairs of the like terms in the above expression –
$ = \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.
$ = \dfrac{1}{{\sqrt 2 }}[0] $
Anything multiplied with zero, gives the resultant value as the zero.
$ = 0 $
Hence, the value of \[\sin (45^\circ + \theta ) - \cos (45^\circ - \theta ) = 0\]
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^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ).
Complete step-by-step answer:
Take the given expression –
\[\sin (45^\circ + \theta ) - \cos (45^\circ - \theta )\]
We can observe that we can apply the standard general formula for $ \sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $ and other identity as $ \cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $ in the above expression.
$ = [\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^\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^\circ = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }} $
$ = [\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-
$ = \dfrac{1}{{\sqrt 2 }}[\cos \theta + \sin \theta - \cos \theta - \sin \theta ] $
Make the pairs of the like terms in the above expression –
$ = \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.
$ = \dfrac{1}{{\sqrt 2 }}[0] $
Anything multiplied with zero, gives the resultant value as the zero.
$ = 0 $
Hence, the value of \[\sin (45^\circ + \theta ) - \cos (45^\circ - \theta ) = 0\]
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^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ).
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