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Geometry Level 3

sin ⁡ 4 ( 1 5 ∘ ) + 4 cos ⁡ 2 ( 1 5 ∘ ) − cos ⁡ 4 ( 1 5 ∘ ) + 4 sin ⁡ 2 ( 1 5 ∘ ) = ? \large \sqrt{\sin^{4}( 15^\circ )+4 \cos^{2} ( 15^\circ ) }-\sqrt{\cos^{4} ( 15^\circ )+ 4 \sin^{2} ( 15^\circ )} = \ ?

2 2 \frac{\sqrt{2}}{2} 2 3 3 \frac{2\sqrt{3}}{3} 3 2 \frac{\sqrt{3}}{2} 3 3 \frac{\sqrt{3}}{3}

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Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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2 solutions

Tanishq Varshney
May 6, 2015

s i n 2 x = 1 − c o s 2 x 2 a n d c o s 2 x = 1 + c o s 2 x 2 sin^{2}x=\frac{1-cos2x}{2}~and~cos^{2}x=\frac{1+cos2x}{2}

here 30 30 means 3 0 o 30^{o} just to avoid typing

( 1 − c o s 30 2 ) 2 + 4 ( 1 + c o s 30 2 ) − ( 1 + c o s 30 2 ) 2 + 4 ( 1 − c o s 30 2 ) \sqrt{(\frac{1-cos30}{2})^{2}+4(\frac{1+cos30}{2})}-\sqrt{(\frac{1+cos30}{2})^{2}+4(\frac{1-cos30}{2})}

after expanding the square and and simplifying

we get ( c o s 30 + 3 2 ) 2 − ( c o s 30 − 3 2 ) 2 \large{\sqrt{(\frac{cos30+3}{2})^{2}}-\sqrt{(\frac{cos30-3}{2})^{2}}}

∣ c o s 30 + 3 2 ∣ − ∣ c o s 30 − 3 2 ∣ \large{|\frac{cos30+3}{2}|-|\frac{cos30-3}{2}|}

c o s 30 + 3 2 − ( 3 − c o s 30 2 ) \large{\frac{cos30+3}{2}-(\frac{3-cos30}{2})}

2 c o s 30 2 = 3 2 \huge{\boxed{\frac{2cos30}{2}=\frac{\sqrt{3}}{2}}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Standard approach. Well done.

Nicely done. I did it a different way but unfortunately ended up getting an incorrect answer, which is really bothering me. Here's how I did it:

Let y= sqrt[((1-cos²(15°))²+4cos²(15°))] - sqrt[(cos²(15°))²+4-4cos²(15°))]

I let u= cos²(15°)

So y= sqrt [(1-u)²+4u] - sqrt[u²-4u+4]

So y= sqrt[u²+2u+1] - sqrt[u²-4u+4]

So y= sqrt[(u+1)²] - sqrt[(u-2)²]

So y= (u+1)-(u-2)

Thus, y=3

If possible, could you please identify the major error that I appear to have made?

Aran Pasupathy - 6 years ago
Aditya Kumar
May 7, 2015

sin ⁡ 4 ( 1 5 ∘ ) + 4 cos ⁡ 2 ( 1 5 ∘ ) \sqrt{\sin^{4}(15^\circ)+4\cos^{2}(15^\circ)} = sin ⁡ 4 ( 1 5 ∘ ) + 4 ( 1 − sin ⁡ 2 ( 1 5 ∘ ) ) \sqrt{\sin^{4}(15^\circ)+4(1-\sin^{2}(15^\circ))} = ( 2 − sin ⁡ 2 ( 1 5 ∘ ) ) 2 \sqrt{(2-\sin^{2}(15^\circ))^2} = 2 − sin ⁡ 2 ( 1 5 ∘ ) 2-\sin^{2}(15^\circ)

Similiarly,

cos ⁡ 4 ( 1 5 ∘ ) + 4 sin ⁡ 2 ( 1 5 ∘ ) \sqrt{\cos^{4}(15^\circ)+4\sin^{2}(15^\circ)} = cos ⁡ 4 ( 1 5 ∘ ) + 4 ( 1 − cos ⁡ 2 ( 1 5 ∘ ) ) \sqrt{\cos^{4}(15^\circ)+4(1-\cos^{2}(15^\circ))} = ( 2 − cos ⁡ 2 ( 1 5 ∘ ) ) 2 \sqrt{(2-\cos^{2}(15^\circ))^2} = 2 − cos ⁡ 2 ( 1 5 ∘ ) 2-\cos^{2}(15^\circ)

The desired difference therefore is: cos ⁡ 2 ( 1 5 ∘ ) − sin ⁡ 2 ( 1 5 ∘ ) \cos^{2}(15^\circ)-\sin^{2}(15^\circ) = cos ⁡ ( 2 × 1 5 ∘ ) \cos(2 \times 15^\circ) = cos ⁡ ( 3 0 ∘ ) \cos(30^\circ) = 3 2 \frac{\sqrt{3}}{2} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

That was unexpected. Superb!

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