Calciferous Cosine

Geometry Level 2

cos ⁡ ( π 4 + x ) + cos ⁡ ( π 4 − x ) = a \large \cos { \left( \frac { \pi }{ 4 } +x \right) } +\cos { \left( \frac { \pi }{ 4 } -x \right) } =a

cos ⁡ ( 3 π 4 + x ) − cos ⁡ ( 3 π 4 − x ) = b \large \cos { \left( \frac { 3\pi }{ 4 } +x \right) } -\cos { \left( \frac { 3\pi }{ 4 } -x \right) } =b

a 2 + b 2 = ? { a }^{ 2 }+{ b }^{ 2 }=\ ?


The answer is 2.

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Shivamani Patil by shivamani patil , 21, India

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4 solutions

Tijmen Veltman
Jun 19, 2015

We can make frequent use of the cosine-sine transformation rule sin ⁡ ( θ + π 2 ) = cos ⁡ ( θ ) \sin\left(\theta+\frac{\pi}2\right)=\cos(\theta) :

a 2 + b 2 = ( cos ⁡ ( π 4 + x ) + cos ⁡ ( π 4 − x ) ) 2 + ( cos ⁡ ( 3 π 4 + x ) − cos ⁡ ( 3 π 4 − x ) ) 2 = ( cos ⁡ ( π 4 + x ) + cos ⁡ ( x − π 4 ) ) 2 + ( cos ⁡ ( 3 π 4 + x ) − cos ⁡ ( x − 3 π 4 ) ) 2 = ( cos ⁡ ( π 4 + x ) + sin ⁡ ( x + π 4 ) ) 2 + ( cos ⁡ ( 3 π 4 + x ) − sin ⁡ ( x − 5 π 4 ) ) 2 = ( cos ⁡ ( π 4 + x ) + sin ⁡ ( x + π 4 ) ) 2 + ( cos ⁡ ( 3 π 4 + x ) − sin ⁡ ( x + 3 π 4 ) ) 2 = cos ⁡ 2 ( π 4 + x ) + sin ⁡ 2 ( π 4 + x ) + cos ⁡ 2 ( 3 π 4 + x ) + sin ⁡ 2 ( 3 π 4 + x ) + 2 sin ⁡ ( π 4 + x ) cos ⁡ ( π 4 + x ) + 2 sin ⁡ ( 3 π 4 + x ) cos ⁡ ( 3 π 4 + x ) = 2 + sin ⁡ 2 ( π 4 + x ) + sin ⁡ 2 ( 3 π 4 + x ) = 2 + sin ⁡ ( 2 x + π 2 ) + sin ⁡ ( 2 x + 3 π 2 ) = 2 + sin ⁡ ( 2 x + π 2 ) − sin ⁡ ( 2 x + π 2 ) = 2 . \begin{aligned} a^2+b^2 &= \left(\cos\left(\frac{\pi}4+x\right)+\cos\left(\frac{\pi}4-x\right)\right)^2\\& +\left(\cos\left(\frac{3\pi}4+x\right)-\cos\left(\frac{3\pi}4-x\right)\right)^2\\ &= \left(\cos\left(\frac{\pi}4+x\right)+\cos\left(x-\frac{\pi}4\right)\right)^2\\& +\left(\cos\left(\frac{3\pi}4+x\right)-\cos\left(x-\frac{3\pi}4\right)\right)^2\\ &= \left(\cos\left(\frac{\pi}4+x\right)+\sin\left(x+\frac{\pi}4\right)\right)^2\\& +\left(\cos\left(\frac{3\pi}4+x\right)-\sin\left(x-\frac{5\pi}4\right)\right)^2\\ &= \left(\cos\left(\frac{\pi}4+x\right)+\sin\left(x+\frac{\pi}4\right)\right)^2\\& +\left(\cos\left(\frac{3\pi}4+x\right)-\sin\left(x+\frac{3\pi}4\right)\right)^2\\ &=\cos^2\left(\frac{\pi}4+x\right)+\sin^2\left(\frac{\pi}4+x\right)\\& +\cos^2\left(\frac{3\pi}4+x\right)+\sin^2\left(\frac{3\pi}4+x\right)\\& +2\sin\left(\frac{\pi}4+x\right)\cos\left(\frac{\pi}4+x\right)\\& +2\sin\left(\frac{3\pi}4+x\right)\cos\left(\frac{3\pi}4+x\right)\\ &=2+\sin 2\left(\frac{\pi}4+x\right)+\sin 2\left(\frac{3\pi}4+x\right)\\ &=2+\sin \left(2x+\frac{\pi}2\right)+\sin \left(2x+\frac{3\pi}2\right)\\ &=2+\sin \left(2x+\frac{\pi}2\right)-\sin \left(2x+\frac{\pi}2\right)\\ &=\boxed{2}.\end{aligned}

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We can use the identities: cos ⁡ ( θ + β ) = cos ⁡ ( θ ) cos ⁡ ( β ) − sin ⁡ ( θ ) sin ⁡ ( β ) \cos(\theta+\beta)=\cos(\theta)\cos(\beta)-\sin(\theta)\sin(\beta) and cos ⁡ ( θ − β ) = cos ⁡ ( θ ) cos ⁡ ( β ) + sin ⁡ ( θ ) sin ⁡ ( β ) \cos(\theta-\beta)=\cos(\theta)\cos(\beta)+\sin(\theta)\sin(\beta) :

a = cos ⁡ ( π 4 ) cos ⁡ ( x ) − sin ⁡ ( π 4 ) sin ⁡ ( x ) + cos ⁡ ( π 4 ) cos ⁡ ( x ) + sin ⁡ ( π 4 ) sin ⁡ ( x ) a = 2 cos ⁡ ( π 4 ) cos ⁡ ( x ) a=\cos\left(\dfrac{\pi}{4}\right)\cos(x)-\sin\left(\dfrac{\pi}{4}\right)\sin(x)+\cos\left(\dfrac{\pi}{4}\right)\cos(x)+\sin\left(\dfrac{\pi}{4}\right)\sin(x) \\ a=2\cos\left(\dfrac{\pi}{4}\right)\cos(x)

b = cos ⁡ ( 3 π 4 ) cos ⁡ ( x ) − sin ⁡ ( 3 π 4 ) sin ⁡ ( x ) − cos ⁡ ( 3 π 4 ) cos ⁡ ( x ) − sin ⁡ ( 3 π 4 ) sin ⁡ ( x ) a = − 2 sin ⁡ ( 3 π 4 ) sin ⁡ ( x ) b=\cos\left(\dfrac{3\pi}{4}\right)\cos(x)-\sin\left(\dfrac{3\pi}{4}\right)\sin(x)-\cos\left(\dfrac{3\pi}{4}\right)\cos(x)-\sin\left(\dfrac{3\pi}{4}\right)\sin(x) \\ a=-2\sin\left(\dfrac{3\pi}{4}\right)\sin(x)

Then we have:

a 2 + b 2 = 4 cos ⁡ 2 ( π 4 ) cos ⁡ 2 ( x ) + 4 sin ⁡ 2 ( 3 π 4 ) sin ⁡ 2 ( x ) = 2 cos ⁡ 2 ( x ) + 2 sin ⁡ 2 ( x ) = 2 \begin{aligned} a^2+b^2&=4\cos^2\left(\dfrac{\pi}{4}\right)\cos^2(x)+4\sin^2\left(\dfrac{3\pi}{4}\right)\sin^2(x) \\ &=2\cos^2(x)+2\sin^2(x) \\ &=\boxed{2} \end{aligned}

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Moderator note:

Wonderful. This is the solution I'm looking for.

Chew-Seong Cheong
Jun 22, 2015

{ a = cos ⁡ ( π 4 + x ) + cos ⁡ ( π 4 − x ) b = cos ⁡ ( 3 π 4 + x ) − cos ⁡ ( 3 π 4 − x ) = − sin ⁡ ( π 4 + x ) + sin ⁡ ( π 4 − x ) \begin{cases} a = \cos{\left(\dfrac{\pi}{4}+x \right)} + \cos{\left(\dfrac{\pi}{4} - x \right)} \\ b = \cos{\left(\dfrac{3\pi}{4}+x \right)} - \cos{\left(\dfrac{3\pi}{4} - x \right)} = - \sin{\left(\dfrac{\pi}{4}+x \right)} + \sin{\left(\dfrac{\pi}{4} - x \right)} \end{cases}

{ a 2 = cos ⁡ 2 ( π 4 + x ) + 2 cos ⁡ ( π 4 + x ) cos ⁡ ( π 4 − x ) + cos ⁡ 2 ( π 4 − x ) b 2 = sin ⁡ 2 ( π 4 + x ) − 2 sin ⁡ ( π 4 + x ) sin ⁡ ( π 4 − x ) + sin ⁡ 2 ( π 4 − x ) \begin{cases} a^2 = \cos^2{\left(\dfrac{\pi}{4}+x \right)} + 2 \cos{\left(\dfrac{ \pi} {4}+x \right)} \cos{\left(\dfrac{\pi}{4} - x \right)} + \cos^2{\left(\dfrac{\pi}{4} - x \right)} \\ b^2 = \sin^2{\left(\dfrac{\pi}{4}+x \right)} - 2 \sin{\left( \dfrac{ \pi} {4}+x \right)} \sin{\left(\dfrac{\pi}{4} - x \right)} + \sin^2{\left(\dfrac{\pi}{4} - x \right)} \end{cases}

⇒ a 2 + b 2 = 1 + 2 cos ⁡ ( π 4 + x ) cos ⁡ ( π 4 − x ) − 2 sin ⁡ ( π 4 + x ) sin ⁡ ( π 4 − x ) + 1 = 1 − 2 cos ⁡ π 2 + 1 = 1 − 0 + 1 = 2 \begin{aligned} \Rightarrow a^2 + b^2 & = 1 + 2 \cos{\left(\dfrac{ \pi} {4}+x \right)} \cos{\left(\dfrac{\pi}{4} - x \right)} \\ & \quad - 2 \sin{\left( \dfrac{ \pi} {4}+x \right)} \sin{\left(\dfrac{\pi}{4} - x \right)} + 1 \\ & = 1 - 2 \cos{\frac{\pi}{2}} + 1 \\ & = 1 - 0 + 1 = \boxed{2} \end{aligned}

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Good solution.Solution which I have in mind is that use cos ⁡ ( a + b ) \cos(a+b) formula to obtain first and second expression as 2 1 / 2 cos ⁡ ( x ) 2^{1/2}\cos(x) and − 2 1 / 2 s i n ( x ) -2^{1/2}sin(x) .Then using sine square plus cos square identity.

shivamani patil - 5 years, 11 months ago

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I thought of that too, but I think this is simpler. Shouldn't it be up-voted?

Chew-Seong Cheong - 5 years, 11 months ago

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Oh I forgot it. Upvoted now :)

shivamani patil - 5 years, 11 months ago
Rooney Great
Jul 19, 2015

just 1+1 =2

remember that cos x = sin (90 - x) :3

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