Show that sin60 - cos45 - - cos60 - sin45 - - -3 - 12-2

L.H.S #10; #10; =sin60^∘cos45^∘+cos #10;60^∘sin45^∘ #10; #10; =√(3)2×1√(2)+12× #10;1√(2) #10; #10; =√(3)2√(2)+12√(2) #10; #10 ...

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Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

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Question

Show that $sin 60^{\circ} cos 45^{\circ} + cos 60^{\circ} sin 45^{\circ} = \frac{\sqrt{3 + 1}}{2 \sqrt{2}}$

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Solution

L.H.S

$= sin 60^{\circ} cos 45^{\circ} + cos 60^{\circ} sin 45^{\circ}$

$= \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}}$

$= \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}$

$= \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Hence, proved.

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Similar questions

Evaluate each of the following

cos $60^{\circ}$ cos $45^{\circ}$ - sin $60^{\circ}$ sin $45^{\circ}$

Show that the value of $\frac{s i n 45^{\circ} - s i n 30^{\circ}}{c o s 45^{\circ} + c o s 60^{\circ}}$ and $\frac{s e c 45^{\circ} - t a n 45^{\circ}}{c o s e c 45^{\circ} + c o t 45^{\circ}}$ are equal

(cos 0^{\circ} + sin 45^{\circ} + sin 30^{\circ}) (sin 90^{\circ} - cos 45^{\circ} + cos 60^{\circ}) =

Q. Evaluate each of the following

cos 60° cos 45° − sin 60° sin 45°

Q. Evaluate the following :

(c o s 0^{\circ} + s i n 45^{\circ} + s i n 30^{\circ}) (s i n 90^{\circ} - c o s 45^{\circ} + c o s 60^{\circ})

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Show that sin60∘cos45∘+cos60∘sin45∘=√3+12√2

L.H.S =sin60∘cos45∘+cos60∘sin45∘ =√32×1√2+12×1√2 =√32√2+12√2 =√3+12√2 Hence, proved.

Show that $sin 60^{\circ} cos 45^{\circ} + cos 60^{\circ} sin 45^{\circ} = \frac{\sqrt{3 + 1}}{2 \sqrt{2}}$

L.H.S

$= sin 60^{\circ} cos 45^{\circ} + cos 60^{\circ} sin 45^{\circ}$

$= \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}}$

$= \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}$

$= \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Hence, proved.