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Find the values of :

(i) 7 sin 30° cos 60°

(ii) 3 sin2 45° + 2 cos2 60°

(iii) cos2 45° + sin2 60° + sin2 30°

(iv) cos 90° + cos2 45° sin 30° tan 45°.

Trigonometrical Ratios

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Answer

(i) 7 sin 30° cos 60°

=7×12×12=74.= 7 \times \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{7}{4}.

Hence, 7 sin 30° cos 60° = 74\dfrac{7}{4}.

(ii) 3 sin2 45° + 2 cos2 60°

=3×(12)2+2×(12)2=3×12+2×14=32+12=42=2.= 3 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 + 2 \times \Big(\dfrac{1}{2}\Big)^2 \\[1em] = 3 \times \dfrac{1}{2} + 2 \times \dfrac{1}{4} \\[1em] = \dfrac{3}{2} + \dfrac{1}{2} \\[1em] = \dfrac{4}{2} \\[1em] = 2.

Hence, 3 sin2 45° + 2 cos2 60° = 2.

(iii) cos2 45° + sin2 60° + sin2 30°

=(12)2+(32)2+(12)2=12+34+14=2+3+14=64=32.= \Big(\dfrac{1}{\sqrt{2}}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{1}{2} + \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{2 + 3 + 1}{4} \\[1em] = \dfrac{6}{4} \\[1em] = \dfrac{3}{2}.

Hence, cos2 45° + sin2 60° + sin2 30° = 32\dfrac{3}{2}.

(iv) cos 90° + cos2 45° sin 30° tan 45°

=0+(12)2×12×1=0+12×12=14.= 0 + \Big(\dfrac{1}{\sqrt{2}}\Big)^2 \times \dfrac{1}{2} \times 1 \\[1em] = 0 + \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{1}{4}.

Hence, cos 90° + cos2 45° sin 30° tan 45° = 14\dfrac{1}{4}.

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