Let α and β be fist quadrant angles with cos(α)= sqrt7/4 and sin(β)= sqrt10/6 Find tan(α-β)

Answer 1

tan(α - β) = (tanα - tanβ)/(1 + tanαtanβ); tanα = √(1 - 7/16)/(√7/4) = 3/√7; tanβ = √10/6/√(1 - 10/36) = √5/13.

tan(α - β) = (3/√7 - √5/13)/(1 + 3/√7·√5/13) =(3√13 - √35)/(√91 + 3√5)

Answer 2

tan(α-β) = sin(α-β) / cos(α-β) = (sinα cosβ - cosα sinβ ) / (cosα cosβ + sinα sinβ) (addition rule)

Since both angles are in the first quadrant, sinα = + sqrt{1 - cos²α) > 0, cosβ = + sqrt(1 - sin²β) > 0, from the Pythagorean identity sin²x + cos²x = 1

Answer 3

Looking at the difference formula for tangent: tan(α-β)=(tan(α)-tan(β))/(1+tan(α)tan(β))

What this formula tells us is that if we know the tangent of α and β, we know the tangent of α-β. Further, we know that tangent=sine/cosine. This means if we can figure out the sine and cosine of both α and β, we can figure out the answer.

The problem says that cos(α)=(√7)/4 and sin(β)=(√10)/6. So, if we can figure out sin(α) and cos(β), we have all the pieces to plug in to the difference formula.

To find sin(α), let's use the fact that cos^2(α)+sin^2(α)=1, or re-arranged: sin(α)=√(1-cos^2(α)) = √(1-7/16)=√(9/16)=3/4. We know to use the positive square root rather than the negative root since the angle is in the first quadrant according to the problem - therefore the sine is positive.

We can use the same trick to find cos(β)=√(1-sin^2(β)) = √(1-10/36)=√(26/36)=(√26)/6.

Now, tan(α)=sin(α)/cos(α) = (3/4)/((√7)/4)=3/√7 = (3√7)/7.

And: tan(β)=sin(β)/cos(β) = ((√10)/6)/((√26)/6)=(√10)/(√26)=(√65)/13.

We now have everything we need to do the difference formula at the top:tan(α-β)=((3√7)/7-(√65)/13)/(1+((3√7)/7)((√65)/13)), which simplifies to (27√(7)-8√(65))/23.