Find the value of sin 2-cos 2- at -30- 60-

The correct option is A 1, 1At θ = 30°, sin 30°=1/2 and cos 30°=√(3)/2. Now, sin^230°+cos^230° =(1/2)^2 + (√(3)/2)^2 =1/4 +3/4 = 1. nbsp; Similarly, at ...

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

Byju's Answer

Standard X

Mathematics

Standard Values of Trigonometric Ratios

Find the valu...

Question

$F i n d t h e v a l u e o f s i n^{2} θ + c o s^{2} θ a t θ = 30^{\circ} & 60^{\circ}$ .

\frac{1}{4}, \frac{3}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{4}, \frac{1}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1, 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0, \frac{3}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1, 1$
$A t θ = 30 °, s i n 30 ° = \frac{1}{2} a n d c o s 30 ° = \frac{\sqrt{3}}{2} . N o w, s i n^{2} 30 ° + c o s^{2} 30 ° = (\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2} = \frac{1}{4} + \frac{3}{4} = 1.$

Similarly,
$a t θ = 60 °, s i n 60 ° = \frac{\sqrt{3}}{2} a n d c o s 30 ° = \frac{1}{2} . N o w, s i n^{2} 60 ° + c o s^{2} 60 ° = (\frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} = \frac{3}{4} + \frac{1}{4} = 1.$

Hence, the value of $s i n^{2} θ + c o s^{2} θ a t θ = 30^{\circ} & 60^{\circ} i s 1.$

Suggest Corrections

Similar questions

Q. Find the value of

s i n^{2} θ + c o s^{2} θ

θ = 30^{\circ} & 60^{\circ}

Q. The value of sin2θ+cos2θ at θ=30∘ & 60∘ is respectively.

cos θ + sin θ = cos 2 θ + sin 2 θ

Find the value of

θ

$If c o t θ = \frac{1}{\sqrt{3}}, find the value of \frac{1 - c o s^{2} θ}{2 - s i n^{2} θ}$

Q. If tan

θ = \frac{12}{13}

, find the value of

\frac{2 sin θ cos θ}{{cos}^{2} θ - {sin}^{2} θ}

Join BYJU'S Learning Program

Find the value of sin2θ+cos2θ at θ=30∘ & 60∘.

The correct option is C 1,1At θ=30°,sin 30°=12 and cos 30°=√32.Now, sin230°+cos230°=(12)2 + (√32)2=14+34=1. Similarly, at θ=60°,sin 60°=√32 and cos 30°=12.Now, sin260°+cos260°=(√32)2 + (12)2=34+14=1. Hence, the value of sin2θ+cos2θ at θ=30∘ & 60∘ is 1.

$F i n d t h e v a l u e o f s i n^{2} θ + c o s^{2} θ a t θ = 30^{\circ} & 60^{\circ}$ .