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bixtya [17]
3 years ago
14

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

os B = 12/13
-220/221
-140/221
140/221
220/221

If A and B are two angles in standard position in Quadrant I, find cos( A -B ) for the given function values. sin A = 3/5 and cos B= 12/37
153/185
57/185
-57/185
-153/185

If A and B are two angles in standard position in Quadrant I, find cos( A - B) for the given function values.
sin A = 15/17 and cos = 3/5
-84/85
-36/85
36/85
84/85

If A and B are two angles in standard position in Quadrant I, find cos( A + B) for the given function values.
sin A = 15/17 and cos = 3/5
-220/221
-140/221
140/221
220/221

If A and B are two angles in standard position in Quadrant I, find cos( A - B) for the given function values.
sin A = 4/5 and cos = 5/13
-33/65
33/65
-63/65
63/65

If A and B are two angles in standard position in Quadrant I, find cos( A + B) for the given function values.
sin A = 3/5 and cos = 12/37
153/185
57/185
-57/185
-153/185
Mathematics
1 answer:
horsena [70]3 years ago
4 0

Answer:

Part 1) cos(A + B) = \frac{140}{221}

Part 2) cos(A - B) = \frac{153}{185}

Part 3) cos(A - B) = \frac{84}{85}

Part 4) cos(A + B) = -\frac{36}{85}

Part 5) cos(A - B) = \frac{63}{65}

Part 6) cos(A+ B) = -\frac{57}{185}

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

sin(A)=\frac{8}{17}

cos(B)=\frac{12}{13}

Determine cos (A+B)

we know that

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{8}{17})^2=1

cos^2(A)+\frac{64}{289}=1

cos^2(A)=1-\frac{64}{289}

cos^2(A)=\frac{225}{289}

cos(A)=\pm\frac{15}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{15}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{13})^2=1

sin^2(B)+\frac{144}{169}=1

sin^2(B)=1-\frac{144}{169}

sin^2(B)=\frac{25}{169}

sin(B)=\pm\frac{25}{169}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{5}{13}

step 3

Find cos(A+B)

substitute in the formula

cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}

cos(A + B) = \frac{180}{221}-\frac{40}{221}

cos(A + B) = \frac{140}{221}

Part 2) we have

sin(A)=\frac{3}{5}

cos(B)=\frac{12}{37}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{3}{5})^2=1

cos^2(A)+\frac{9}{25}=1

cos^2(A)=1-\frac{9}{25}

cos^2(A)=\frac{16}{25}

cos(A)=\pm\frac{4}{5}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{4}{5}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{37})^2=1

sin^2(B)+\frac{144}{1,369}=1

sin^2(B)=1-\frac{144}{1,369}

sin^2(B)=\frac{1,225}{1,369}

sin(B)=\pm\frac{35}{37}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{35}{37}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}

cos(A - B) = \frac{48}{185}+\frac{105}{185}

cos(A - B) = \frac{153}{185}

Part 3) we have

sin(A)=\frac{15}{17}

cos(B)=\frac{3}{5}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}

cos(A - B) = \frac{24}{85}+\frac{60}{85}

cos(A - B) = \frac{84}{85}

Part 4) we have

sin(A)=\frac{15}{17}        

cos(B)=\frac{3}{5}

Determine cos (A+B)

we know that    

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}      

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A+B)

substitute in the formula    

cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}

cos(A + B) = \frac{24}{85}-\frac{60}{85}

cos(A + B) = -\frac{36}{85}

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