Prove that-isin 60- cos30-cos60- sin30-1ii sin4 -7-7 cos-9-7-cos4 -9-7 sin-9-7-3-2iiisin 3 -8-5 cos-8-5-cos3 -8-5 sin-8-5-1

The correct option is (i)sin(60^∘ θ)cos(30^∘+θ)+cos(60^∘ θ)sin(30^∘+ θ) =sin[(60^∘ θ)+(30^∘+θ)] =sin[60^∘ θ+30^∘+θ] =sin(90^∘) =1 =RHS ∴ LHS=RHS Hence prove ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Common Angles

Prove that:is...

Question

Prove that:
(i) $s i n (60^{\circ} - θ) c o s (30^{\circ} + θ) + c o s (60^{\circ} - θ) s i n (30^{\circ} + θ) = 1$
(ii) $s i n (\frac{4 π}{7} + 7) c o s (\frac{π}{9} + 7) - c o s (\frac{4 π}{9} + 7) s i n (\frac{π}{9} + 7) = \frac{\sqrt{3}}{2}$
(iii) $s i n (\frac{3 π}{8} - 5) c o s (\frac{π}{8} + 5) + c o s (\frac{3 π}{8} - 5) s i n (\frac{π}{8} + 5) = 1$

Open in App

Solution

(i) $s i n (60^{\circ} - θ) c o s (30^{\circ} + θ) + c o s (60^{\circ} - θ) s i n (30^{\circ} + θ)$
$= s i n [(60^{\circ} - θ) + (30^{\circ} + θ)]$
$= s i n [60^{\circ} - θ + 30^{\circ} + θ]$
$= s i n (90^{\circ})$
=1
=RHS
$∴ L H S = R H S$
Hence proved.
(ii)LHS: $s i n (\frac{4 π}{7} + 7) c o s (\frac{π}{9} + 7) - c o s (\frac{4 π}{9} + 7) s i n (\frac{π}{9} + 7)$
= $s i n (\frac{4 π}{9} - \frac{π}{9})$
$= s i n \frac{3 π}{9} = s i n \frac{π}{3} = \frac{\sqrt{3}}{2}$
=RHS
$∴ L H S = R H S$
Hence proved.
(iii) $s i n (\frac{3 π}{8} - 5) c o s (\frac{π}{8} + 5) + c o s (\frac{3 π}{8} - 5) s i n (\frac{π}{8} + 5)$
$s i n [(\frac{3 π}{8} - 5) + (\frac{π}{8} + 5)]$
$= s i n (\frac{3 π}{8} + \frac{π}{8})$
$= s i n \frac{4 π}{8} = s i n \frac{π}{2} = 1$
=RHS
$∴$ LHS=RHS
Hence proved.

Suggest Corrections

Similar questions

Q. Prove that:
(i) sin (60° − θ) cos (30° + θ) + cos (60° − θ) sin (30° + θ) = 1.
(ii)

\sin (\frac{4 π}{9} + 7) \cos (\frac{π}{9} + 7) - \cos (\frac{4 π}{9} + 7) \sin (\frac{π}{9} + 7) = \frac{\sqrt{3}}{2}

(iii)

\sin (\frac{3 π}{8} - 5) \cos (\frac{π}{8} + 5) + \cos (\frac{3 π}{8} - 5) \sin (\frac{π}{8} + 5) = 1

Q. Which of the following will come next in the following series?

919829873987649876559876546987654

Q. Show that:

(- \frac{8}{9} \times \frac{1}{- 5}) + (- \frac{8}{9} \times - \frac{- 7}{11}) = - \frac{8}{9} \times (\frac{1}{- 5} + \frac{- 7}{11})

Fill in the blanks:

$(i) - 4 \times \frac{7}{9} = \frac{7}{9} \times$ ___

$(i i) \frac{5}{11} \times \frac{- 3}{8} = \frac{- 3}{8} \times$ ___

$(i i i) \frac{1}{2} \times (\frac{3}{4} + \frac{- 5}{12}) = \frac{1}{2} \times$ ____ $+ (\times) \frac{- 5}{12}$

$(i v) \frac{- 4}{5} \times \frac{5}{7} \times$ ___ $= (\frac{- 4}{5} \times \frac{5}{7}) \times \frac{- 8}{9}$

Q. Study the following number line and answer the questions that follow.
2 5 9 1 7 2 5 1 9 7 3 9 5 2 4 6 8 1 9 7 5 2 7 3 1 9 7 8 5 2 6 5 4 1 9 7 8 2 0 1 0 9 7 8 5 1 9 7 3 5 1 6 9 7 2 1 7 3 7 9 5 1

How many times is number 5 followed by 1 or 2, but not preceded by 8 ?

Join BYJU'S Learning Program

Prove that: (i)sin(60∘−θ)cos(30∘+θ)+cos(60∘−θ)sin(30∘+θ)=1 (ii) sin(4π7+7)cos(π9+7)−cos(4π9+7)sin(π9+7)=√32 (iii)sin(3π8−5)cos(π8+5)+cos(3π8−5)sin(π8+5)=1