John asked in Science & MathematicsMathematics · 1 month ago

# Let sin A = − 5/13 with A in QIII and find then following; tan 2A?

I have tried to solve this and I do not understand how to. Please explain how you got the answer. Thank you

Update:

The question has been answered, thank you all!

Relevance
• TomV
Lv 7
1 month ago

Both the sine and the cosine of an angle in Q3 are negative.

If A is in Q3, and sinA = -5/13, cosA = -√(1 - 25/169) = - 12/13

tanA = sinA/cosA = (-5)/(-12) = 5/12

The double angle formula for the tangent of an angle is:

tan 2A = 2tanA/(1-tan²A)

Ans:

tan 2A = 2(5/12)/(1-25/144)

= 10*144/[12(144-25)]

= 120/119

===============================================

A = 202.62°

[arctan(5/12) = 22.62, but since A is in Q3, A = arctan(5/12) + 180°]

2A = 45.24°

[2*202.62 = 405.24°, but since that is greater the 360°, subtract 360 to express the angle in the interval [0, 360°). ]

tan(2A) = tan(45.24°) = 1.0084 or 120/119 (to 5 sig. dig.)

• 1 month ago

The Trig. Identity for Tan (2A) is 2Tan(A) / (1 - Tan^2(A))

Also Tan (A) = Sin(A) / Cos(A)

Since Sin(A) = -5/13

Then by Pythagoras Cos(A)  = -12/13  (Quadrant (III)

Hence Tan(A) = (-5/13)/(-12/13) = 5/12

NB The '-' and '13' cancel out.

Hence Tan(2A) = 2(5/12) / ( 1 - (5/12)^2)

Tan(2A) = (10/12) / (1 - 25/144)

Tan(2A) = (10/12)/(119/144)

Tan(2A) = 120/144)/(119/144)

Tan(2A) = 120/119    (NB The '144' cancels out)

Tan(2A) = 1.008403361...

Tan(2A) ~ 1.0084

A ~ 112.61986... degrees.

• Philip
Lv 6
1 month ago

A is in (pi , 3pi/2). Therefore 2A is in (2pi , 3pi) which is co-terminal with (0 , pi).

sinA = (-5/13). cosA =-[1- sin^2(A)]^(1/2) =-[1-(-5/13)^2]^(1/2)= -[(169-25)/169]^(1/2)

= (-12/13). Then tanA = sinA/cosA = (5/12).

Now tan(2A) = 2tanA/[1 - tan^2(A)] = 2(5/12)/[1 - (5/12)^2] = (5/6)/[(144-25)/144] =

(120/119). Now arctan(120/119) = 45.23972990°. Clearly 2A is co-terminal with this

angle and is, therefore, in (0 , pi/2).

• 1 month ago

sinA = -5/13...in Q3

so, tanA = 5/12...also in Q3 and positive

Now, tan2A = 2tanA/(1 - tan²A)

Then, tan2A = 2(5/12)/(1 - (5/12)²)

i.e. (5/6)/(119/144) = 120/119

:)>