Tanalpha=3/4,secalpha=13/5 neither aplha beta is in 1st quadrant find cos(alpha-beta)?

Question

Caroline Adams · Accepted Answer

Assuming you meant sec beta =13/5
#color(white)("XXX")cos(alpha-beta)=-56/65#

If #tan(alpha)=3/4# and #alpha# is not in Quadrant I
then #alpha# is an angle in Quadrant III (based on the CAST system, a positive value fro #tan# only occurs in Quadrants I and III.

Similarly, if #sec(beta)color(white)("xx")[=1/(cos(beta))]color(white)("xx")=13/5# and #beta# is not in Quadrant I
then #beta# is an angle in Quadrant IV.

We can draw these angles with appropriate right angled triangles and using the Pythagorean Theorem determine the length of the third side of each triangle (in the image below the calculated side's length is the circled value).

Using the #cos# angle-difference formula:
#color(white)("XXX")cos(alpha-beta)=cos(alpha)cos(beta) + sin(alpha)sin(beta)#

We can use the above images to evaluate this as
#color(white)("XXX")cos(alpha-beta)=((-4)/5) * (5/12) + ((-3)/5) *(12/13)#

#color(white)("XXX"cos(alpha-beta))=(-20)/65+ (-36)/65 = -56/65#

HIX Tutor · Answer

undefined To find $ \cos(\alpha - \beta) $, we can use the formula:

$$
\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta
$$

We have $ \tan \alpha = \frac{3}{4} $ and $ \sec \alpha = \frac{13}{5} $.

First, let's find $ \cos \alpha $ and $ \sin \alpha $ for $ \alpha $:

1. **Find $ \cos \alpha $**:
   $$
   \sec \alpha = \frac{1}{\cos \alpha}
   $$
   $$
   \cos \alpha = \frac{5}{13}
   $$

2. **Find $ \sin \alpha $** using $ \tan \alpha $:
   $$
   \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3}{4}
   $$
   So, 
   $$
   \sin \alpha = \tan \alpha \times \cos \alpha = \frac{3}{4} \times \frac{5}{13} = \frac{15}{52}
   $$

Now we need to find $ \beta $ using $ \sec \beta = \frac{5}{13} $ too. Since $ \sec \beta = \frac{13}{5} $, we can find $ \beta $:

3. **Find $ \cos \beta $**:
   $$
   \sec \beta = \frac{1}{\cos \beta}
   $$
   $$
   \cos \beta = \frac{5}{13}
   $$

4. **Find $ \sin \beta $** using $ \tan \beta = \frac{3}{4} $:
   $$
   \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{3}{4}
   $$
   So,
   $$
   \sin \beta = \tan \beta \times \cos \beta = \frac{3}{4} \times \frac{5}{13} = \frac{15}{52}
   $$

Now, we can substitute into the cosine difference formula.

5. **Substitute values**:
   $$
   \cos(\alpha - \beta) = \left(\frac{5}{13} \times \frac{5}{13}\right) + \left(\frac{15}{52} \times \frac{15}{52}\right)
   $$
   $$
   = \frac{25}{169} + \frac{225}{2704}
   $$

To add these, we need a common denominator:

6. **Find common denominator**:
   The common denominator of $ 169 $ and $ 2704 $ is $ 2704 $.

Convert $ \frac{25}{169} $:
   $$
   \frac{25 \times 16}{169 \times 16} = \frac{400}{2704}
   $$

Now, add:
$$
\cos(\alpha - \beta) = \frac{400}{2704} + \frac{225}{2704} = \frac{625}{2704}
$$

So, the answer is:
$$
\cos(\alpha - \beta) = \frac{625}{2704}
$$