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Question Number 208282 by hardmath last updated on 10/Jun/24

If cosα−cosβ = (1/5) sinα + sinβ = (1/2) Find cos(α + β) = ?

$$\mathrm{If}\:\:\:\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{sin}\alpha\:+\:\mathrm{sin}\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{cos}\left(\alpha\:+\:\beta\right)\:=\:? \\ $$

Commented by mr W last updated on 12/Jun/24

what you wrote means cosα−cosβ =(1/2) (1/5) sinα + sinβ = (1/2) but actually you meant cosα−cosβ = (1/5) sinα + sinβ = (1/2) they are very different questions! it′s misleading! can you please pay more attention?

$${what}\:{you}\:{wrote}\:{means} \\ $$$$\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{sin}\alpha\:+\:\mathrm{sin}\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${but}\:{actually}\:{you}\:{meant} \\ $$$$\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{sin}\alpha\:+\:\mathrm{sin}\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${they}\:{are}\:{very}\:{different}\:{questions}! \\ $$$${it}'{s}\:{misleading}! \\ $$$${can}\:{you}\:{please}\:{pay}\:{more}\:{attention}? \\ $$

Commented by Frix last updated on 10/Jun/24

I solved both versions but one gives no useable exact result, thus I think the other one was meant.

$$\mathrm{I}\:\mathrm{solved}\:\mathrm{both}\:\mathrm{versions}\:\mathrm{but}\:\mathrm{one}\:\mathrm{gives}\:\mathrm{no} \\ $$$$\mathrm{useable}\:\mathrm{exact}\:\mathrm{result},\:\mathrm{thus}\:\mathrm{I}\:\mathrm{think}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{one}\:\mathrm{was}\:\mathrm{meant}. \\ $$

Answered by Frix last updated on 10/Jun/24

cos (α+β) =cos α cos β −sin α sin β (cos α −cos β =(1/5))^2 ⇒ cos α cos β =((cos^2 α +cos^2 β)/2)−(1/(50)) (sin α +sin β =(1/2))^2 ⇒ sin α sin β =(1/8)−((sin^2 α +sin^2 β)/2) ⇒ cos (α+β) =((171)/(200))

$$\mathrm{cos}\:\left(\alpha+\beta\right)\:=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\left(\mathrm{cos}\:\alpha\:−\mathrm{cos}\:\beta\:=\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\:\:\:\:\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:=\frac{\mathrm{cos}^{\mathrm{2}} \:\alpha\:+\mathrm{cos}^{\mathrm{2}} \:\beta}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{50}} \\ $$$$\left(\mathrm{sin}\:\alpha\:+\mathrm{sin}\:\beta\:=\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\:\:\:\:\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:=\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha\:+\mathrm{sin}^{\mathrm{2}} \:\beta}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)\:=\frac{\mathrm{171}}{\mathrm{200}} \\ $$

Commented by efronzo1 last updated on 10/Jun/24

cos α−cos β=(1/2)

$$\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by hardmath last updated on 10/Jun/24

yes dear professor thank you

$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by hardmath last updated on 10/Jun/24

answer: 171/200

$$\mathrm{answer}:\:\mathrm{171}/\mathrm{200} \\ $$

Commented by efronzo1 last updated on 10/Jun/24

$$\:\underbrace{\:} \\ $$

Commented by Frix last updated on 10/Jun/24

Then solve it if you can!

$$\mathrm{Then}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{if}\:\mathrm{you}\:\mathrm{can}! \\ $$

Commented by efronzo1 last updated on 11/Jun/24

read the original question cos α−cos β=(1/5)sin α+sin β=(1/2) but your cos α−cos β=(1/5) and sin α+sin β=(1/2)

$$\mathrm{read}\:\mathrm{the}\:\mathrm{original}\:\mathrm{question} \\ $$$$\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{but}\:\mathrm{your}\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\mathrm{and}\:\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 11/Jun/24

the confusion was caused by hardmath sir through his misleading writing of the question.

$${the}\:{confusion}\:{was}\:{caused}\:{by}\: \\ $$$$\boldsymbol{{hardmath}}\:{sir}\:{through}\:{his}\: \\ $$$${misleading}\:{writing}\:{of}\:{the}\:{question}. \\ $$

Commented by Frix last updated on 11/Jun/24

@efronzo1: As I wrote before I also solved this one but it gives no useable exact result, we can only approximate it. So I decided that the other version with the "nice" result was the right one.

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