How do you use product to sum formulas to write the product $6 {sin 15}^{\circ} {sin 45}^{\circ}$ $6sin15^circsin45^circ$ as a sum or difference?

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Answer 1 · 2017-01-08T14:39:08

$6 {sin 15}^{\circ} {sin 45}^{\circ} = 3 \sqrt{3} - 3$ $6sin15^@sin45^@=3sqrt3-3$

As $cos (A - B) - cos (A + B) = 2 sin A sin B$ $cos(A-B)-cos(A+B)=2sinAsinB$

$6 {sin 15}^{\circ} {sin 45}^{\circ} = 6 {sin 45}^{\circ} {sin 15}^{\circ}$ $6sin15^@sin45^@=6sin45^@sin15^@$

= $6 cos (45^{\circ} - 15^{\circ}) - 6 cos (45^{\circ} + 15^{\circ})$ $6cos(45^@-15^@)-6cos(45^@+15^@)$

= $6 {cos 30}^{\circ} - 6 {cos 60}^{\circ}$ $6cos30^@-6cos60^@$

= $6 \times \frac{\sqrt{3}}{2} - 6 \times \frac{1}{2}$ $6xxsqrt3/2-6xx1/2$

= $3 \sqrt{3} - 3$ $3sqrt3-3$

How do you use product to sum formulas to write the product 6sin15∘sin45∘6sin15^circsin45^circ as a sum or difference?