cos−1310+cos−125=? Let cos−1310=α,cos−125=β then, cos⁡α=310,cos⁡β=25 Therefore cos⁡α=3⋅2225=322⋅cos⁡β

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cos−1310+cos−125=?  Let cos−1310=α,cos−125=β then, cos⁡α=310,cos⁡β=25  Therefore  cos⁡α=3⋅2225=322⋅cos⁡β

sonorous9n · Accepted Answer

Use trig identity:sin2θ+cos2θ=1  ⇒sin⁡α=1−cos2α=1−(310)2=110∀ 0≤α≤π2  ⇒sin⁡β=1−cos2β=1−(25)2=15∀ 0≤β≤π2  Now, use trig identity  cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β   cos⁡(α+β)=31025−11015=12  ⇒α+β=cos−112=π4   (∴cos⁡(α+β)∈[−1,1])  ∴cos−1310+cos−125=π4  Or alternatively use trig identity  sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β  sin⁡(α+β)=11025+31015=12  ⇒α+β=sin−112=π4   (∵sin⁡(α+β)∈[−1,1])  ∴cos−1310+cos−125=π4

Kirsten Davis · Accepted Answer

Hint : apply this formula:  cos−1x+cos−1y=cos−1[xy−(1−x2)(1−y2)]  Put x=210 and y=25

Vasquez · Accepted Answer

Like Proof for the formula of sum of arcsine functions arcsin⁡x+arcsin⁡y using cos⁡(A+B) and the definition of principal values cos−1x+cos−1y={cos−1(xy−(1−x2)(1−y2))  if  cos−1x+cos−1y≤π2π−cos−1(xy−(1−x2)(1−y2))  otherwise   Now  cos−1⁡x+cos−1⁡y≤πwill happen⇔cos−1⁡x≤π−cos−1⁡y=cos−1⁡(−y)⇔π2−sin−1⁡x≤π2−sin−1⁡(−y)⇔sin−1⁡x≥sin−1⁡(−y)⇔x≥−y

\cos^{-1} \frac{3}{\sqrt{10}}+\cos^{-1} \frac{2}{\sqrt5}=? Let \cos^{-1} \frac{3}{\sqrt{10}}=\alpha, \cos^{-1} \frac{2}{\sqrt5}=\beta then, \cos \alpha=\frac{3}{\sqrt{10}},

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