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Question: Find \[\sin 74^\circ \] using \[\sin (A + B) = \sin A\cos B + \cos A\sin B\] Find \[\sin 16^\circ...

Find sin⁡74∘\sin 74^\circ using sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B
Find sin⁡16∘\sin 16^\circ using sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B

Explanation

Solution

Here we have to find the values of the given trigonometric ratios using the given trigonometric identities. We can find the trigonometric ratios using the sine and cosine table. Sine and cosine table gives us the trigonometric values at different angles. Trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles.

Formula used:
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B

Complete step by step solution:
We have to find sin⁡74∘\sin 74^\circ .
Now sin⁡74∘\sin 74^\circ can be written as sin⁡(37∘+37∘)\sin (37^\circ + 37^\circ ) .
⇒sin⁡(74∘)=sin⁡(37∘+37∘)\Rightarrow \sin \left( {74^\circ } \right) = \sin \left( {37^\circ + 37^\circ } \right)
By using the trigonometric identity, sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B in above equation, we get
⇒sin⁡(74∘)=sin⁡37∘cos⁡37∘+cos⁡37∘sin⁡37∘\Rightarrow \sin \left( {74^\circ } \right) = \sin 37^\circ \cos 37^\circ + \cos 37^\circ \sin 37^\circ
Adding the terms, we have
⇒sin⁡(74∘)=2sin⁡37∘cos⁡37∘\Rightarrow \sin \left( {74^\circ } \right) = 2\sin 37^\circ \cos 37^\circ
From the sine and cosine table, we have
⇒sin⁡(74∘)=2×0.6018×0.7986\Rightarrow \sin \left( {74^\circ } \right) = 2 \times 0.6018 \times 0.7986
By multiplying the terms, we get
⇒sin⁡(74∘)=0.9713\Rightarrow \sin \left( {74^\circ } \right) = 0.9713
Now we have to find sin⁡16∘\sin 16^\circ .
sin⁡16∘\sin 16^\circ can be written as sin⁡(30∘−14∘)\sin (30^\circ - 14^\circ )
⇒sin⁡(16∘)=sin⁡(30∘−14∘)\Rightarrow \sin \left( {16^\circ } \right) = \sin \left( {30^\circ - 14^\circ } \right)
By using the trigonometric identity sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B in the above equation, we get
⇒sin⁡(16∘)=sin⁡30∘cos⁡14∘−cos⁡30∘sin⁡14∘\Rightarrow \sin \left( {16^\circ } \right) = \sin 30^\circ \cos 14^\circ - \cos 30^\circ \sin 14^\circ
From the sine and cosine table, we have
⇒sin⁡(16∘)=0.5×0.9703−0.8660×0.2419\Rightarrow \sin \left( {16^\circ } \right) = 0.5 \times 0.9703 - 0.8660 \times 0.2419
By multiplying the terms, we have
⇒sin⁡(16∘)=0.48515−0.2095\Rightarrow \sin \left( {16^\circ } \right) = 0.48515 - 0.2095
Subtracting the terms, we get
⇒sin⁡(16∘)=0.27565\Rightarrow \sin \left( {16^\circ } \right) = 0.27565

Therefore, sin⁡(74∘)=0.9713\sin \left( {74^\circ } \right) = 0.9713 and sin⁡(16∘)=0.27565\sin \left( {16^\circ } \right) = 0.27565.

Note:
We can find the value of sine and cosine using the table. Usually sine tables will accompany other tables of trigonometric functions. Hyperbolic trigonometric functions are logarithmic and are not natural sine, cosines or tangents. We can observe that the table of natural sine and natural cosines are generally divided into the following parts. They are the following:
(i) In the extreme left, vertical column of the table ,the angles are from 0∘0^\circ to 90∘90^\circ at intervals of 1∘1^\circ .
(ii) In another vertical column, about the middle of the table, the angles are from 89∘89^\circ to 0∘0^\circ at intervals of 1∘1^\circ .
(iii) In the horizontal row at the top of the table, the angles are from 0′0' to 60′60' at intervals of 10′10'.
(iv) In the horizontal row at the bottom of the table, the angles are from 60′60' to 0′0' at intervals of 10′10'.
(v) In the horizontal row at the extreme right of the table, the angles are from 1′1' to 9′9' at intervals of 1′1'. This part of the table is known as Mean Difference Column.