What are the tangent sum and difference identities?
tan(A + B) = (tan A + tan B)/(1 − tan A·tan B); tan(A − B) = (tan A − tan B)/(1 + tan A·tan B).
Video Summary
Tangent sum: tan(A+B) = (tanA + tanB)/(1 - tanA·tanB); difference: tan(A−B) = (tanA − tanB)/(1 + tanA·tanB).
Proof uses tan = sin/cos and the sine/cosine sum identities.
Example: tan(−105°) = tan(−60° − 45°) → simplifies and rationalizes to 2 + √3.
Example: tan(5π/12) = tan(π/4 + π/6) → simplifies to 2 + √3 after algebra and rationalization.
Identity: tan(π − θ) = −tan(θ); unit circle explains the sign change.
tan(A + B) = (tan A + tan B)/(1 − tan A·tan B); tan(A − B) = (tan A − tan B)/(1 + tan A·tan B).
Write −105° as −60° − 45°, apply the difference identity, substitute tan values (tan(−60°)=−√3, tan45°=1), simplify and rationalize to get 2 + √3.
Convert 5π/12 to 75° and use tan(π/4 + π/6): substitute tan(π/4)=1 and tan(π/6)=1/√3, then simplify and rationalize to obtain 2 + √3.
On the unit circle tan = y/x; tan(π) = 0 and the sum/difference identity shows the numerator becomes −tanθ while the denominator stays 1, giving −tanθ. The unit circle sign change explains this.
The goal of the video is to use the sum and difference identities to determine function values.
The lesson focuses on the sum and difference identities specifically for tangent functions. These identities help in calculating the tangent of the sum or difference of two angles.
For the sum of two angles, the tangent is calculated using the formula:
[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}.
]
Conversely, for the difference of two angles, the formula is:
[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}.
]
The presentation highlights that while these formulas look similar, the operations in the numerator and the denominator differ. The sum formula has addition in the numerator and subtraction in the denominator, while the difference formula has subtraction in the numerator and addition in the denominator.
The first step is remembering that tangent theta equals sine theta divided by cosine theta.
The proof of the tangent sum identity relies on the definitions of sine and cosine combined with the sum identities for these functions.
While the proof is not detailed in the video, it emphasizes the need to understand that tangent is derived from the ratio of sine and cosine.
Viewers are encouraged to look over the necessary derivations before moving on to application problems.
To determine the exact value of tangent -105 degrees, use reference angles to add or subtract them to obtain -105 degrees.
The example demonstrates how to find the tangent of -105 degrees by identifying reference angles of -60 and -45 degrees.
The difference identity is used here, setting up the equation:
[
\tan(-60^\circ - 45^\circ) = \frac{\tan(-60^\circ) - \tan(45^\circ)}{1 + \tan(-60^\circ) \cdot \tan(45^\circ)}.
]
The video then guides through calculating the tangent values and simplifying the resulting expression step-by-step.
The tangent of -105 degrees, however, is often presented in a rationalized form.
After finding the tangent of -105 degrees as a fraction, instructions are given on how to rationalize the denominator by multiplying by the conjugate.
This simplification process involves additional algebra, ultimately leading to a more manageable expression for tangent -105 degrees.
The numerator and denominator are simplified systematically to achieve the final exact value.
Convert this to degrees first, which simplifies the reference angle process.
The tangent problem is reiterated by converting radians to degrees, where 5π/12 translates to 75 degrees.
Two reference angles of 45 degrees and 30 degrees are identified for this problem, using the sum formula for tangent:
[
\tan\left(\frac{\pi}{4} + \frac{\pi}{6}\right).
]
The identity is applied, and calculations for tangent values follow similarly to previous examples.
When we apply this identity for tangent π - theta, we find that it equals negative tangent theta.
In this final segment, the video applies an identity to express tangent π - theta as a single function of theta.
The calculation shows that:
[
\tan(\pi - \theta) = -\tan(\theta).
]
This is visually explained using the unit circle, demonstrating the relationship effectively.