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Geometry Level 2

sin ⁡ 10 5 ∘ + cos ⁡ 10 5 ∘ = ? \sin 105 ^ \circ +\cos 105^ \circ =?

2 \sqrt{2} 2 1 2 \frac{1} { \sqrt{2} } 1

Istiak Reza by Istiak Reza , 23, Bangladesh

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8 solutions

Mahindra Bhoi
Dec 8, 2014

(sin105+cos105)^2=1+ 2sin105cos105=1+sin210=1-sin30=1/2

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Shahbaz Ali
Nov 30, 2014

sin(105) + cos(105)

= sin(60 + 45 ) + cos( 60 + 45 )

= sin(60)cos(45) + cos(60)sin(45) + cos(60)cos(45) - sin(60)sin(45)

= sqrt3/2 * 1/ sqrt2 + 1/2* 1/ sqrt2 + 1/2* 1/ sqrt2 - sqrt3/2 * 1/ sqrt2

first and fourth terms cancel each other and adding 2nd and 3rd term

= 1/ sqrt2

4 Helpful 0 Interesting 0 Brilliant 0 Confused

(sin105+cos105)^2=1+ 2sin105cos105=1+sin210=1-sin30=1/2

Ngố Xác - 6 years, 6 months ago

I used the same method. Retyping it in LaTex for you.

sin ⁡ 10 5 ∘ + cos ⁡ 10 5 ∘ = sin ⁡ ( 6 0 ∘ + 4 5 ∘ ) + cos ⁡ ( 6 0 ∘ + 4 5 ∘ ) \sin {105^\circ}+\cos{105^\circ} = \sin {(60^\circ + 45^\circ) } + \cos { ( 60 ^ \circ + 45 ^ \circ ) }

= sin ⁡ ( 6 0 ∘ + 4 5 ∘ ) + cos ⁡ ( 6 0 ∘ + 4 5 ∘ ) = sin ⁡ 6 0 ∘ cos ⁡ 4 5 ∘ + cos ⁡ 6 0 ∘ sin ⁡ 4 5 ∘ =\sin {(60^\circ + 45^\circ) } + \cos {(60 ^\circ + 45^\circ)} = \sin {60^\circ } \cos { 45^\circ } + \cos {60^\circ } \sin { 45^\circ }

= ( 3 2 ) ( 1 2 ) + ( 1 2 ) ( 1 2 ) + ( 1 2 ) ( 1 2 ) − ( 3 2 ) ( 1 2 ) = 1 2 = (\frac {\sqrt{3}} {2} )( \frac {1} {\sqrt{2}}) + (\frac {1} {2} )( \frac {1}{\sqrt{2}}) + (\frac {1} {2} )( \frac {1}{\sqrt{2}} ) - (\frac {\sqrt{3}} {2} )( \frac {1} {\sqrt{2}}) = \boxed {\frac {1}{\sqrt{2}}}

Chew-Seong Cheong - 6 years, 6 months ago
Kathleen Kasper
Dec 8, 2014

Let\quad sin105°+cos105°=x <hr>Then(sin105°+cos105°)²=x²\ sin²105°+2sin105°cos105°+cos²105°=x²\ 1+sin(2*105°)=x²\ 1+sin210°=x²\ 1/2=x²\ x=+-sqrt(1/2)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Let's use the identity sin ⁡ a + cos ⁡ a = 2 sin ⁡ ( a + 4 5 ∘ ) \sin a+\cos a=\sqrt{2}\sin(a+45^\circ) :

sin ⁡ 10 5 ∘ + cos ⁡ 10 5 ∘ = 2 sin ⁡ 15 0 ∘ = 1 2 \sin 105^\circ + \cos 105^\circ=\sqrt{2}\sin 150^\circ=\frac{1}{\sqrt{2}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

sin105 + cos105 = sin(15 + 90) + cos(15+90) = cos(15) - sin(15) = 1/sqrt2

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Farouk Yasser
Dec 8, 2014

sin(105) + cos(105) = x

(sin(105) + cos(105))^2 = x^2

sin^2(105) + 2cos(105)sin(105) + cos^2(105) = x^2 (1)

.....................

sin^2(105) + cos^2(105) = 1

2cos(105)sin(105) = sin(210) = -sin(30)

Writing 1 again becomes:

1 - sin(30) = X^2

1 - 1/2 = x^2

x = 1/rooot 2

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Jitesh Mittal
Dec 8, 2014

We can do it directly just by looking at options:

Sin 105 lies between 0 to 1

Cos 105 lies between 0 to -1

=> Sin105+Cos105 lies between -1 to 1

Only one option satisfies it so answer becomes 1/ sqrt2

We can also do it by calculating sin 105 and cos 105 and then adding them as shown in other solutions.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Istiak Reza
Dec 1, 2014

There are many ways to solve it...... sin 105°+ cos 105° =sin ( 90°+ l5°) + cos 1O5° = cos 15°+ cos 105° = 2 cos( ( 105° +15°)/ 2) cos( ( 105°-l5°) / 2) = 2 cos 6o° cos45° = 2X 1/2 x 1 / sqrt 2 = 1 / sqrt 2

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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